Indefinite Integration - PDF Free Download

Indefinite Integration 1 An antiderivative of a function y = f(x) defined on some interval (a, b) is called any function F(x) whose derivative at any point of this interval is equal to f(x): F'(x) = f(x) If F(x) is an antiderivative of f(x), then the function of the form F(x) + C, where C is an arbitrary constant, is also an antiderivative of f(x) 2 The indefinite integral of a function f(x) is the collection of all antiderivatives for this function: 3 The derivative of the indefinite integral is equal to the integrand: 4 The indefinite integral of the sum of two functions is equal to the sum of the integrals: 5 The indefinite integral of the difference of two functions is equal to the difference of the integrals: 6 A constant factor can be moved across the integral sign: 7 8 9 10 Page 1 of 90

11 Integration by substitution 12 Integration by parts where u(x), v(x) are differentiable functions Definition of the Antiderivative and Indefinite Integral The function F(x) is called an antiderivative of f(x), if The family of all antiderivatives of a function f(x) is called the indefinite integral of the function f(x) and is denoted by Thus, if F is a particular antiderivative, we may write where C is an arbitrary constant Properties of the Indefinite Integral In formulas given below f and g are functions of the variable x, F is an antiderivative of f, and a, k, C are constants Page 2 of 90

Table of Integrals It's supposed below that a, p (p 1), C are real constants, b is the base of the exponential function (b 1, b > 0) Page 3 of 90

Q1 Calculate Q2 Calculate the integral Transforming the integrand and using the formula for integral of the power function, we have Q3 Calculate We use the table integral Then Q4 Find the integral Using the table integral we obtain, Page 4 of 90

Q5 Calculate the integral Since, the integral is Q6 Find the integral Using the double angle formula sin 2x = 2 sin x cos x and the identity sin 2 x + cos 2 x = 1, we can write: Q7 Evaluate the integral Decompose the integrand into partial functions: Equate coefficients: Hence, Page 5 of 90

Then The integral is equal to Q8 Evaluate First we divide the numerator by the denominator, obtaining Then Q9 Evaluate the integral We can write: Q10 Evaluate the integral Decompose the integrand into partial functions: Page 6 of 90

Equate coefficients: Hence, Then The integral is equal to Q11Evaluate Decompose the integrand into the sum of two fractions: Equate coefficients: Hence Page 7 of 90

The integrand can be written as The initial integral becomes Q12 Find the integral We can factor the denominator in the integrand: Decompose the integrand into partial functions: Equate coefficients: Hence, Then Page 8 of 90

Now we can calculate the initial integral: Q13 Calculate the integral Rewrite the denominator in the integrand as follows: The factors in the denominator are irreducible quadratic factors since they have no real roots Then Equate coefficients: Page 9 of 90

This yields Hence, Integrating term by term, we obtain the answer: We can simplify this answer Let Then Page 10 of 90

Hence, The complete answer is Q14 Evaluate the integral We can factor the denominator in the integrand: Decompose the integrand into partial functions: Equate coefficients: Hence, Thus, the integrand becomes Page 11 of 90

So, the complete answer is Q15Calculate the integral Decompose the integrand into partial functions, taking into account that the denominator has a third degree root: Equate coefficients: We get the following system of equations: Hence, Page 12 of 90

The initial integral is equal to Q16 Find the integral Since is reducible, we complete the square in the denominator: Now, we can compute the integral using the reduction formula Then Q17 Find the integral We make the substitution: Page 13 of 90

Then Q18 Calculate the integral We make the following substitution: Then the integral (we denote it by I ) becomes Divide the fraction: As a result, we have Q19 Evaluate the integral We can write the integral as Since the least common multiple (LCM) of the denominators of the fractional powers is equal to n = LCM(1,3) = 3, we make the substitution: Page 14 of 90

Thus Make the new substitution: The final answer is Q20 Evaluate the integral We can write the integral as Make the substitution: The integral becomes Since the degree of the numerator is greater than the degree of the denominator, we divide the numerator by the denominator: The integral is equal to Page 15 of 90

Q21 Find the integral We can write the integral as As can be seen, the least common multiple (LCM) of the denominators of the fractional powers is equal to n = LCM(3,4) = 12, we make the substitution: This yields The degree of the numerator is greater than the degree of the denominator, therefore, we divide the fraction: After simple transformations we obtain the final answer: Q22 Evaluate Make the substitution: Page 16 of 90

This yields Q23Calculate the integral We make the substitution As a result, we have Q24 Evaluate the integral We use the universal trigonometric substitution: Since we have, Q25 Calculate the integral Page 17 of 90

Make the universal trigonometric substitution: Then the integral becomes Q26 Find the integral As in the previous examples, we will use the universal trigonometric substitution: Since we can write: Q27 Evaluate We can write the integral in form: Page 18 of 90

Use the universal trigonometric substitution: This leads to the following result: Q28 Calculate the integral Since, we can write Hence, so the integral can be transformed in the following way: Make the substitution Then use the identity We get the final answer: Page 19 of 90

Q29 Calculate the integral We solve this integral by making the trigonometric substitution Taking into account that we find the integral: Q30 Find the integral We make the substitution: Page 20 of 90

As a result, the integral becomes Using the method of partial fractions, we can write the integrand as Calculate the coefficients A, B and C: Hence, So the integral is Integration of Some Special Classes of Trigonometric Functions In this section we consider 8 classes of integrals with trigonometric functions Special transformations and subtitutions used for each of these classes allow us to obtain exact solutions for these integrals 1 Integrals of the form To find integrals of this type, use the following trigonometric identities: Page 21 of 90

2 Integrals of the form It's assumed here and below that m and n are positive integers To find an integral of this form, use the following substitutions: a If the power n of the cosine is odd (the power m of the sine can be arbitrary), then the substitution is used b If the power m of the sine is odd, then the substitution is used c If both powers m and n are even, then first use the double angle formulas to reduce the power of the sine or cosine in the integrand Then, if necessary, apply the rules a) or b) 3 Integrals of the form The power of the integrand can be reduced by using the trigonometric identity and the reduction formula 4 Integrals of the form We can reduce the power of the integrand using the trigonometric identity Page 22 of 90

and the reduction formula 5 Integrals of the form This type of integrals can be simplified with help of the reduction formula: 6 Integrals of the form Similarly to the previous examples, this type of integrals can be simplified by the formula 7 Integrals of the form a If the power of the secant n is even, then using the identity the secant function is expressed asthe tangent function The factor is separated and used for transformation of the differential As a result,the entire integral (including differential) is expressed as the function of tan x b If both the powers n and m are odd, then the factor sec x tan x, which is necessary to transform the differential, is separated Then the entire integral is expressed through sec x c If the power of the secant n is odd, and the power of the tangent m is even, then the tangent is d expressed as the secant using the identity Then the integrals of the secant are calculated 8 Integrals of the form a If the power of the cosecant n is even, then using the identity cosecant function is expressed as the cotangent function The factor is separated and used for transformation of the differential As a result, the integrand and differential are expressed through cot x Page 23 of 90

b If both the powers n and m are odd, then the factor csc x cot x, which is necessary to transform the differential, is separated Then the integral is expressed through csc x c If the power of the cosecant n is odd, and the power of the cotangent m is even, then the cotangent is expressed as the cosecant using the identity Then the integrals of the cosecant are calculated Q31 Calculate the integral Let u = cos x, du = sin xdx Then Q32 Evaluate the integral Making the substitution u = sin x, du = cos xdx and using the identity, we obtain Q33 Find the integral Using identities and, we can write: Page 24 of 90

Calculate the integrals in the latter expression To find the integral, we make the substitution u = sin 2x, du = 2cos 2xdx Then Hence, the initial integral is Q34 Calculate the integral We can write: Transform the integrand using the identities We get Q35 Evaluate the integral Page 25 of 90

Making the substitution u = cos x, du = sin xdx and expressing the sine through cosine with help of the formula, we obtain Q36 Evaluate the integral Transform the integrand by the formula Hence, Then the integral becomes Q37 Evaluate the integral We use the identity to transform the integral This yields Page 26 of 90

Q38 Calculate the integral Using the identity, we have Q39 Calculate the integral We use the reduction formula Hence, The integral is a table integral which is equal to (It can be easily found usingthe universal trigonometric substitution ) As a result, the integral becomes Q40 Evaluate the integral We use the reduction formula Page 27 of 90

Hence, Q41Compute Q42 Compute Use the identity Then Since and is a table integral equal to, we obtain the following complete answer: Page 28 of 90

Q43 Evaluate the integral We make the substitution: Then Q44 Calculate the integral To find the integral, we make the substitution: Using the identity we have, Express sin t in terms of x: Hence, the integral is Page 29 of 90

Q45 Calculate the integral We make the substitution: Hence, Then the integral becomes Returning back to the variable x : we obtain the complete answer: Q46 Evaluate the integral We make the trigonometric substitution: x = a sec t, dx = a tan t sec t dt Calculate the integral using the identity : For t, we have the following expression: Page 30 of 90

Hence, returning back to the variable x, we obtain: Q47 Evaluate the integral We can write the integral as Make the trigonometric substitution: Now we can calculate the integral: Q48 Calculate the integral First we complete the square in the integrand Then, making the substitution and using the identity we can write:, Page 31 of 90

Q49 Find the integral Make the substitution: Then Express and in terms of x: Hence, Page 32 of 90

Q50 Evaluate the integral Using the trigonometric substitution we get Now, we make the substitution The integral becomes Use the method of partial fractions to transform the integrand: Equate the coefficients: Then Hence, we can write the integrand as So, the initial integral is equal to Page 33 of 90

Returning back to the variable x, we have Hence, the final answer is Page 34 of 90

The Definite Integral Let f (x) be a continuous function on the closed interval [a, b] The definite integral of f (x) from a to b is defined to be the limit where THE LIMIT DEFINITION OF A DEFINITE INTEGRAL The following problems involve the limit definition of the definite integral of a continuous function of one variable on a closed, bounded interval Begin with a continuous function on the interval Let be an arbitrary (randomly selected) partition of the interval into subintervals (subdivisions) Let, which divides the interval be the sampling numbers (or sampling points) selected from the subintervals That is, is in, is in, is in,, Page 35 of 90

is in, and is in, is in Define the mesh of the partition to be the length of the largest subinterval That is, let for and define The definite integral of on the interval is most generally defined to be For convenience of computation, a special case of the above definition uses subintervals of equal length and sampling points chosen to be the right-hand endpoints of the subintervals Thus, each subinterval has length equation for and the right-hand endpoint formula is equation for The definite integral of on the interval can now be alternatively defined by Page 36 of 90

We will need the following well-known summation rules 1 (n times), where is a constant 2 3 4 5, where is a constant 6 PROBLEM 1 : Use the limit definition of definite integral to evaluate PROBLEM 2 : Use the limit definition of definite integral to evaluate PROBLEM 3 : Use the limit definition of definite integral to evaluate PROBLEM 4 : Use the limit definition of definite integral to evaluate PROBLEM 5 : Use the limit definition of definite integral to evaluate Page 37 of 90

PROBLEM 6 : Use the limit definition of definite integral to evaluate PROBLEM 7 : Use the limit definition of definite integral to evaluate SOLUTIONS TO THE LIMIT DEFINITION OF A DEFINITE INTEGRAL SOLUTION 1 : Divide the interval into equal parts each of length for by Choose the sampling points to be the right-hand endpoints of the subintervals and for The function is Then the definite integral is Page 38 of 90

(Since is the variable of the summation, the expression is a constant Use summation rule 1 from the beginning of this section) SOLUTION 2 : Divide the interval into equal parts each of length for Choose the sampling points to be the right-hand endpoints of the subintervals and given by for The function is Then the definite integral is Page 39 of 90

(Use summation rule 6 from the beginning of this section) (Use summation rules 5 and 1 from the beginning of this section) (Use summation rule 2 from the beginning of this section) Page 40 of 90

SOLUTION 3 : Divide the interval into equal parts each of length for Choose the sampling points to be the right-hand endpoints of the subintervals and given by for The function is Then the definite integral is Page 41 of 90

(Use summation rule 6 from the beginning of this section) (Use summation rules 1 and 5 from the beginning of this section) (Use summation rule 2 from the beginning of this section) SOLUTION 4 : Divide the interval into equal parts each of length for Choose the sampling points to be the right-hand endpoints of the subintervals and given by Page 42 of 90

for The function is Then the definite integral is (Use summation rule 6 from the beginning of this section) (Use summation rules 5 and 1 from the beginning of this section) (Use summation rule 2 from the beginning of this section) Page 43 of 90

SOLUTION 5 : Divide the interval into equal parts each of length for Choose the sampling points to be the right-hand endpoints of the subintervals and given by for The function is Then the definite integral is Page 44 of 90

(Use summation rule 6 from the beginning of this section) (Use summation rules 5 and 1 from the beginning of this section) (Use summation rule 2 from the beginning of this section) Page 45 of 90

SOLUTION 6 : Divide the interval into equal parts each of length for by Choose the sampling points to be the right-hand endpoints of the subintervals and for The function is Then the definite integral is Page 46 of 90

(Use summation rule 6 from the beginning of this section) (Use summation rules 1 and 5 from the beginning of this section) (Use summation rules 2 and 3 from the beginning of this section) SOLUTION 7 : Divide the interval into equal parts each of length for by Choose the sampling points to be the right-hand endpoints of the subintervals and Page 47 of 90

for The function is Then the definite integral is (Use summation rule 5 from the beginning of this section) (Use summation rule 4 from the beginning of this section) Page 48 of 90

Properties of the Definite Integral We assume below that f (x) and g (x) are continuous functions on the closed interval [a, b] 1 2 where k is a constant; 3 4 5 If for all, then 6 7 Page 49 of 90

8 If in the interval [a, b], then The Fundamental Theorem of Calculus Let f (x) be a function, which is continuous on the closed interval [a, b] If F (x) is any antiderivative of f (x) on [a, b],then The Area under a Curve The area under the graph of the function f (x) between the vertical lines x = a, x = b (Figure 1) is given by the formula Fig1 Fig2 Let F (x) and G (x) be indefinite integrals of functions f (x) and g (x), respectively If f (x) g (x) on the closed interval [a, b], then the area between the curves y = f (x), y = g (x) and the lines x = a, x = b (Figure 2) is given by The Method of Substitution for Definite Integrals The definite integral Page 50 of 90

of the variable x can be changed into an integral with respect to t by making the substitution x = g (t): The new limits of integration for the variable t are given by the formulas where g -1 is the inverse function to g, ie t = g -1 (x) Integration by Parts for Definite Integrals In this case the formula for integration by parts looks as follows: where means the difference between the product of functions uv at x = b and x = a Q1 Evaluate the integral Using the fundamental theorem of calculus, we have Q2 Calculate the integral Q3Evaluate the integral Page 51 of 90

First we make the substitution: Determine the new limits of integration When x = 0, then t = 1 When x = 1, then we have t = 2 So, the integral with the new variable t can be easily calculated: Q4 Evaluate the integral We can write Apply integration by parts: In this case, let Hence, the integral is Page 52 of 90

Q5 Find the area bounded by the curves and First we find the points of intersection (see Figure 3): Fig3 As can be seen, the curves intercept at the points (0,0) and (1,1) Hence, the area is given by Q6 Find the area bounded by the curves and First we find the points of intersection of the curves (Figure 4): The upper boundary of the region is the parabola and the lower boundary is the straight line Page 53 of 90

The area is given by Q8 Find the area inside the ellipse By symmetry (see Figure 6), the area of the ellipse is twice the area above the x-axis The latter is given by To calculate the last integral, we use the trigonometric substitution x = asin t, dx = acos tdt Refine the limits of integration When x = a, then sin t = 1, and When x = a, then sin t = 1, Thus we get Page 54 of 90

Hence, the total area of the ellipse is πab Try to Your Self Q 1 Find the area lying above the x-axis and included between the circle x 2 +y 2 =8x and the parabola y 2 =4x Q 2 Find the area lying above the x-axis and included between the circle x 2 +y 2 =16a2 and the parabola y 2 =6ax Q 3 Find the area of the smaller region bounded by the ellipse and the line Q 4 Find the area of the region included between x 2 =4y, y = 2, y = 4 and the y-axis in the first quadrant Q 5 Find the area between the parabolas 4ay = x 2 and y 2 = 4ax Q 6 Find the area bounded by the curve y 2 = 4ax and the line y = 2a and y-axis Q 7 Find the area bounded by the parabola y2 = 8x and its latus rectum Q 8 Find the area of the circle x 2 + y 2 = 16, which is exterior to the parabola y 2 = 6x Q 9 Sketch the region common to the circle x 2 + y 2 = 8 and the parabola x 2 = 4y Also find the area of the common region using integration Q 10 Draw the rough sketch of the region and find the area enclosed by the region using method of integration Q 11 Using integration, find the area of the triangle ABC whose vertices are A(2,3), B(2,8) and C(6,5) Q 12 Using integration, find the area of the triangle ABC whose vertices are A(2,5), B(4,7) and C(6,2) Page 55 of 90

Q 13 Using integration, find the area of the triangle ABC whose vertices are A(-1,1), B(05) and C(3,2) Q 14 Compute the area bounded by the lines x+2y = 2, y-x=1 and 2x+y = 7 Q 15 Compute the area bounded by the lines y = 4x+5, y = 5-x, and 4y = x+5 Q 16 Compute the area bounded by the lines 2x+y = 4, 3x-2y = 6, and x-3y+5=0 Q 17 Using integration, find the area of the region bounded by x-7y+19=0, and y =çxú Q 18 Using integration, find the area of the region bounded by the line i 2y= -x+8, x-axis and the lines x = 2 and x = 4 ii y -1 = x, x-axis and the lines x = -2 and x = 3 iii y =, line y = x and the positive x- axis Q 19 Find the area of the region enclosed between the two circles x 2 + y 2 = 1 and (x-4) 2 + y 2 =16 Q 20 Find the area of the region in the first quadrant enclosed by the x-axis, the line y = 4x and the circle x 2 + y 2 = 32 Q 21 Find the area of the smaller part of the circle x 2 + y 2 =a 2 cut off by the line Q 22 Find the area of the region Q 23 Sketch the graph of the curve y = and evaluate Q 24 Sketch the graph of the curve and find the area bounded by y =, x=-2, x=3, y=0 Q 25 Find the area bounded by the line y = sin2x and y = cos2x between x = 0 and x=p/4 Q 26 Prove that the curves y2 = 4x and x2 = 4y divide the area of the square bounded by x = 0, x = 4, y = 4 and y = 0 into three equal parts Page 56 of 90

Differential Equation Separable Equations A first order differential equation y' = f(x,y) is called a separable equation if the function f(x,y) can be factored into the product of two functions of x and y: where p(x) and h(y) are continuous functions Considering the derivative y' as the ratio of two differentials and divide the equation by h(y): we move dx to the right side Of course, we need to make sure that h(y) 0 If there's a number x 0 such that h(x 0 ) = 0, then this number will also be a solution of the differential equation Division by h(y) causes loss of this solution Denoting, we write the equation in the form We have separated the variables so now we can integrate this equation: where C is an integration constant Calculating the integrals, we get the expression representing the general solution of the separable differential equation Q1 Solve the differential equation In the given case p(x) = 1 and h(y) = y(y +2) Divide the equation by h(y) and move dx to the right side: One can notice that by dividing we can lose the solutions y = 0 and y = 2 when h(y) becomes zero Page 57 of 90

In fact, let's see that y = 0 is a solution of the differential equation Obviously, Substituting this into the equation gives 0 = 0 Hence, y = 0 is one of the solutions Similarly, we can check thaty = 2 is also a solution Returning to the differential equation, we integrate it: We can calculate the left integral by using the fractional decomposition of the integrand: Thus, we get the following decomposition of the rational integrand: Hence, We can rename the constant: 2C = C 1 Thus, the final solution of the equation is written in the form Page 58 of 90

Here the general solution is expressed in implicit form In the given case we can transform the expression to obtain the answer as an explicit function y = f(x,c 1 ), where C 1 is a constant However, it's possible to do not for all differential equations Q2 Solve the differential equation We can rewrite this equation in the following way: Divide both sides by (x 2 + 4)y to get Obviously, that x 2 + 4 0 for all real x Check if y = 0 is a solution of the equation Substituting y = 0 and dy = 0 into the differential equation, we see that the function y = 0 is one of the solutions of the equation Now we can integrate it: Notice that dx 2 = d(x 2 + 4) Hence, We can represent the constant C as lnc 1, where C 1 > 0 Then Thus, the given differential equation has the following solutions: This answer can be simplified Indeed, if using an arbitrary constant C, which takes values from to +, the solution can be written in the form: Page 59 of 90

When C = 0, it becomes y = 0 Q3 Find all solutions of the differential equation y' = xe y Transform this equation in following way: Obviously, that division by e y does not lead to the loss of solutions, since e y > 0 After integrating we have This answer can be expressed in the explicit form: We assume in the latter expression that the constant C > 0 in order to satisfy the domain of the logarithmic function Q4 Find a particular solution of the differential equation under condition y(1) = 1 Divide both sides of the equation by x: We suppose that x 0, because the domain of the given equation is x > 0 Integrating this equation yields: Page 60 of 90

The integral in the right side is calculated as follows: Hence, the general solution in the implicit form is given by where C 1 = 2C is an integration constant Find the values of C 1 to satisfy the initial condition y(1) = 1: Thus, the particular solution satisfying the initial condition is written in the following way: Q5 Solve the differential equation y' cot 2 x + tan 2 y = 0 We write this equation as follows: Divide both sides of the equation by tan y cot 2 x: Check for possible missed solutions when dividing There might be the following two roots: Substituting this into the initial equations, we see that The second possible solution is given by is a solution Page 61 of 90

Here we get the answer which does not satisfy the initial differential equation Now we can integrate the given equation and find its general solution: The final answer is given by Q6 Find a particular solution of the equation satisfying the initial condition y(0) = 0 Write this equation in the following way: Divide both sides by 1 + e x : Since 1 + e x > 0, then we did not miss solutions of the original equation Integrating this equation yields: Page 62 of 90

Determine the constant C from the initial condition y(0) = 0 Hence, the final answer is given by Q7 Solve the equation The product xy in each side does not allow separating the variables Therefore, we make the replacement: The relationship for differential is given by Substituting this into the equation, we can write: By multiplying both sides by x and then canceling the corresponding fractions in the left and right side, we get Take into account that x = 0 is a solution of the equation, which can be checked by direct substitution Simplify the latter expression: Page 63 of 90

Now the variables x and t are separated: After integrating we have By making the reverse substitution t = xy, we find the general solution of the equation: The complete answer is written in the form: Homogeneous Equations Definition of Homogeneous Differential Equation A first order differential equation is called homogeneous equation, if the right side satisfies the condition for all t In other words, the right side is a homogeneous function (with respect to the variables x and y) of the zero order: A homogeneous differental equation can be also written in the form or alternatively, in the differential form: Page 64 of 90

where P(x,y) and Q(x,y) are homogeneous functions of the same degree Definition of Homogeneous Function A function P(x,y) is called a homogeneous function of the degree n if the following relationship is valid for all t > 0: Solving Homogeneous Differential Equations A homogeneous equation can be solved by substitution y = ux, which leads to a separable differential equation Q1 Solve the differential equation It is easy to see that the polynomials P(x,y) and Q(x,y), respectively, at dx and dy are homogeneous functions of the first order Therefore, the original differential equation is also homogeneous Suppose that y = ux, where u is a new function depending on x Then Substituting this into the differential equation, we obtain Hence, Dividing both sides by x yields: When dividing by x, we could lose the solution x = 0 The direct substitution shows that x = 0 is indeed a solution of the given differential equation Integrate the latter expression to obtain: where C is a constant of integration Returning to the old variable y, we can write: Thus, the equation has two solutions: Page 65 of 90

Q2 Solve the differential equation We notice that the root x = 0 does not belong to the domain of the differential equation Rewrite the equation in the form: As seen, this equation is homogeneous Make the substitution y = ux Hence, Substituting this expression into the equation gives: Divide by x 0 to get: We obtain the separable equation: The next step is to integrate the left and the right side of the equation: Hence, Here the constant C can be written as ln C1 (C1 > 0) Then Page 66 of 90

Thus, we have got two solutions: If C1 = 0, the answer is y = xe and we can make sure that it is also a solution to the equation Indeed, substituting into the differential equation, we obtain: Then all the solutions can be represented by one formula: where C is an arbitrary real number Q3 Solve the differential equation Here we deal with a homogeneous equation In fact, we can rewrite it in the form: Make the substitution y = ux Then y' = u'x + u Substituting y and y' into the original equation, we have Divide both sides by ux2 We notice that x =0 is not the solution of the equation However, one can check that u =0or y =0 is one of the solutions of the differential equation As a result, we have Page 67 of 90

Integrating, we find the general solution: Taking into account that, we can write the last expression in the form The given expression can be represented as an explicit inverse function x(y): Since C is an arbitrary real number, we can replace the "minus" sign before the constant to the "plus" sign Then Thus, the given differential equation has the solutions: Example 4 Solve the differential equation As it follows from the right side of the equation, x 0 and y 0 We can make the substitution y = ux, y' = u'x + u This yields: Page 68 of 90

Integrating this equation, we obtain: Let the constant 2C be denoted by just C Hence, Thus, the general solution is written in the form Q5 Find the general solution of the differential equation It is easy to see that the given equation is homogeneous Therefore, we can use the substitution y = ux, y' = u'x + u As a result, the equation is converted into the separable differential equation: Divide both sides by x3 (Notice that x = 0 is not the solution) Now we can integrate the last equation: Since u = y/x, the solution can be written in the form: Page 69 of 90

It follows from here that We can denote ec = C1, (C1 > 0) Then the solution in the implicit form is given by the equation where the constant C1 > 0 Linear Differential Equations of First Order Definition of Linear Equation of First Order A differental equation of type where a(x) and f(x) are continuous functions of x, is called a linear nonhomogeneous differential equation of first order We consider two methods of solving linear differential equations of first order: Using an integrating factor; Using an Integrating Factor If a linear differential equation is written in the standard form: the integrating factor is defined by the formula Multiplying the left side of the equation by the integrating factor u(x) converts the left side into the derivative of the product y(x)u(x) The general solution of the differential equation is expressed as follows: where C is an arbitrary constant Method of Variation of a Constant This method is similar to the previous approach First it's necessary to find the general solution of Page 70 of 90

the homogeneous equation: The general solution of the homogeneous equation contains a constant of integration C We replace the constant C with a certain (still unknown) function C(x) By substituting this solution into the nonhomogeneous differential equation, we can determine the function C(x) The described algorithm is called the method of variation of a constant Of course, both methods lead to the same solution Initial Value Problem If besides the differential equation, there is also an initial condition in the form of y(x0) = y0, such a problem is called the initial value problem (IVP) or Cauchy problem A particular solution for an IVP does not contain the constant C, which is defined by substitution of the general solution into the initial condition y(x0) = y0 Q1 Solve the equation y' y xex = 0 Rewrite this equation in standard form: We will solve this equation using the integrating factor Then the general solution of the linear equation is given by Q2 Solve the differential equation We will solve this problem by using the method of variation of a constant First we find the general solution of the homogeneous equation: which is solved by separating the variables: Page 71 of 90

where C is a positive real number Now we replace C with a certain (still unknown) function C(x) and will find a solution of the original nonhomogeneous equation in the form: Then the derivative is given by Substituting this into the equation gives: Upon integration, we find the function C(x): where C1 is an arbitrary real number Thus, the general solution of the given equation is written in the form Q3 Solve the equation y' 2y = x First we solve this problem using an integrating factor The given equation is already written in the standard form Therefore Then the integrating factor is The general solution of the original differential equation has the form: Page 72 of 90

We calculate the last integral with help of integration by parts Then Q4 Solve the differential equation x2y' + xy + 2 = 0 We solve this problem using the method of variation of a constant For convenience, we write this equation in the standard form: Divide both sides by x 2 Obviously, that x = 0 is not the solution of the equation Consider the homogeneous equation: After easy transformations we find the answer y = C/x, where C is any real number The last expression includes the case y = 0, which is also a solution of the homogeneous equation Now we replace the constant C with the function C(x) and substitute the solution y = C(x)/x into the initial nonhomogeneous differential equation Since we obtain Page 73 of 90

Thus, the general solution of the initial equation is given by Q5 Solve the initial value problem: First of all we calculate the integrating factor, which is written as Here Hence, the integrating factor is given by We can take the function u(x) = cos x as the integrating factor One can make sure that the left side of the equation is the derivative of the product y(x)u(x): Then the general solution of the given equation is written in the form: Next we determine the value of C, which satisfies the initial condition y(0) = 1: It follows from here that C = 4/3 Hence, the solution for the initial value problem is given by Page 74 of 90

Q6 Solve the differential equation (IVP) with the initial condition y(1) = 2 Determine the integrating factor: We can take the function u(x) = x3 as the integrating factor One can check that the left side of the equation is derivative of the product y(x)u(x): The general solution of the differential equation is written as Now we can find the constant C using the initial condition y(1) = 2 Substituting the general solution into this condition gives: Thus, the solution of the IVP is given by Q7 Find the general solution of the differential equation y = (2y 4 + 2x)y' It can be seen that this equation is not linear with respect to the function y(x) However, we can try to find the solution for the inverse function x(y) We write the given equation through differentials and make some transformations: Now we see that we have a linear differential equation with respect to the function x(y) We can solve it with help of the integrating factor: Page 75 of 90

Then the general solution as the inverse function x(y) is given by Linear Differential Equations of First Order Definition of Linear Equation of First Order A differental equation of type where a(x) and f(x) are continuous functions of x, is called a linear nonhomogeneous differential equation of order We consider two methods of solving linear differential equations of first order: Using an integrating factor; Method of variation of a constant Using an Integrating Factor If a linear differential equation is written in the standard form: the integrating factor is defined by the formula Multiplying the left side of the equation by the integrating factor u(x) converts the left side into the derivative product y(x)u(x) The general solution of the differential equation is expressed as follows: where C is an arbitrary constant Method of Variation of a Constant This method is similar to the previous approach First it's necessary to find the general solution of the homoge equation: The general solution of the homogeneous equation contains a constant of integration C We replace the constant Cwith a certain (still unknown) function C(x) By substituting this solution into the nonhomogeneou differential equation, we can determine the function C(x) The described algorithm is called the method of variation of a constant Of course, both methods lead to the s solution Initial Value Problem If besides the differential equation, there is also an initial condition in the form of y(x0) = y0, such a problem called the initial value problem (IVP) or Cauchy problem A particular solution for an IVP does not contain the constant C, which is defined by substitution of the gener solution into the initial condition y(x0) = y0 Example 1 Page 76 of 90

Solve the equation y' y xex = 0 Rewrite this equation in standard form: We will solve this equation using the integrating factor Then the general solution of the linear equation is given by Example 2 Solve the differential equation We will solve this problem by using the method of variation of a constant First we find the general solution o homogeneous equation: which is solved by separating the variables: where C is a positive real number Now we replace C with a certain (still unknown) function C(x) and will find a solution of the original nonhomogeneous equation in the form: Then the derivative is given by Substituting this into the equation gives: Page 77 of 90

Upon integration, we find the function C(x): where C1 is an arbitrary real number Thus, the general solution of the given equation is written in the form Example 3 Solve the equation y' 2y = x A First we solve this problem using an integrating factor The given equation is already written in the standar Therefore Then the integrating factor is The general solution of the original differential equation has the form: We calculate the last integral with help of integration by parts Then B Now we construct the solution by the method of variation of a constant Consider the corresponding homo equation: and find its general solution where C again denotes any real number Notice that at C = 0, we get y = 0 that is also a solution of the homog equation Page 78 of 90

Next we suppose that C is a function of x and substitute the solution y = C(x)e2x into the initial nonhomogen equation We can write Hence, This integral was already found above in section A, so we obtain As a result, the general solution of the nonhomogeneous differential equation is given by As you can see, both methods give the same answer :) Example 4 Solve the differential equation x2y' + xy + 2 = 0 We solve this problem using the method of variation of a constant For convenience, we write this equation in standard form: Divide both sides by x2 Obviously, that x = 0 is not the solution of the equation Consider the homogeneous equation: After easy transformations we find the answer y = C/x, where C is any real number The last expression inclu case y = 0, which is also a solution of the homogeneous equation Now we replace the constant C with the function C(x) and substitute the solution y = C(x)/x into the initial nonhomogeneous differential equation Since we obtain Page 79 of 90

Thus, the general solution of the initial equation is given by Example 5 Solve the initial value problem: First of all we calculate the integrating factor, which is written as Here Hence, the integrating factor is given by We can take the function u(x) = cos x as the integrating factor One can make sure that the left side of the equ the derivative of the product y(x)u(x): Then the general solution of the given equation is written in the form: Next we determine the value of C, which satisfies the initial condition y(0) = 1: It follows from here that C = 4/3 Hence, the solution for the initial value problem is given by Example 6 Page 80 of 90

Solve the differential equation (IVP) with the initial condition y(1) = 2 Determine the integrating factor: We can take the function u(x) = x3 as the integrating factor One can check that the left side of the equation is derivative of the product y(x)u(x): The general solution of the differential equation is written as Now we can find the constant C using the initial condition y(1) = 2 Substituting the general solution into this condition gives: Thus, the solution of the IVP is given by Example 7 Find the general solution of the differential equation y = (2y4 + 2x)y' It can be seen that this equation is not linear with respect to the function y(x) However, we can try to find the solution for the inverse function x(y) We write the given equation through differentials and make some transformations: Now we see that we have a linear differential equation with respect to the function x(y) We can solve it with the integrating factor: Then the general solution as the inverse function x(y) is given by Page 81 of 90

Linear Differential Equations of First Order Definition of Linear Equation of First Order A differental equation of type where a(x) and f(x) are continuous functions of x, is called a linear nonhomogeneous differential equation of order We consider two methods of solving linear differential equations of first order: Using an integrating factor; Method of variation of a constant Using an Integrating Factor If a linear differential equation is written in the standard form: the integrating factor is defined by the formula Multiplying the left side of the equation by the integrating factor u(x) converts the left side into the derivative product y(x)u(x) The general solution of the differential equation is expressed as follows: where C is an arbitrary constant Method of Variation of a Constant This method is similar to the previous approach First it's necessary to find the general solution of the homoge equation: The general solution of the homogeneous equation contains a constant of integration C We replace the constant Cwith a certain (still unknown) function C(x) By substituting this solution into the nonhomogeneou differential equation, we can determine the function C(x) The described algorithm is called the method of variation of a constant Of course, both methods lead to the s solution Initial Value Problem If besides the differential equation, there is also an initial condition in the form of y(x0) = y0, such a problem called the initial value problem (IVP) or Cauchy problem A particular solution for an IVP does not contain the constant C, which is defined by substitution of the gener solution into the initial condition y(x0) = y0 Example 1 Page 82 of 90

where C1 is an arbitrary real number Thus, the general solution of the given equation is written in the form Example 3 Solve the equation y' 2y = x A First we solve this problem using an integrating factor The given equation is already written in the standar Therefore Then the integrating factor is The general solution of the original differential equation has the form: We calculate the last integral with help of integration by parts Then B Now we construct the solution by the method of variation of a constant Consider the corresponding homo equation: and find its general solution where C again denotes any real number Notice that at C = 0, we get y = 0 that is also a solution of the homog equation Next we suppose that C is a function of x and substitute the solution y = C(x)e2x into the initial nonhomogen Page 84 of 90

equation We can write Hence, This integral was already found above in section A, so we obtain As a result, the general solution of the nonhomogeneous differential equation is given by As you can see, both methods give the same answer :) Example 4 Solve the differential equation x2y' + xy + 2 = 0 We solve this problem using the method of variation of a constant For convenience, we write this equation in standard form: Divide both sides by x2 Obviously, that x = 0 is not the solution of the equation Consider the homogeneous equation: After easy transformations we find the answer y = C/x, where C is any real number The last expression inclu case y = 0, which is also a solution of the homogeneous equation Now we replace the constant C with the function C(x) and substitute the solution y = C(x)/x into the initial nonhomogeneous differential equation Since we obtain Page 85 of 90

Solve the differential equation (IVP) with the initial condition y(1) = 2 Determine the integrating factor: We can take the function u(x) = x3 as the integrating factor One can check that the left side of the equation is the derivative of the product y(x)u(x): The general solution of the differential equation is written as Now we can find the constant C using the initial condition y(1) = 2 Substituting the general solution into this condition gives: Thus, the solution of the IVP is given by Example 7 Find the general solution of the differential equation y = (2y4 + 2x)y' It can be seen that this equation is not linear with respect to the function y(x) However, we can try to find the solution for the inverse function x(y) We write the given equation through differentials and make some transformations: Now we see that we have a linear differential equation with respect to the function x(y) We can solve it with the integrating factor: Then the general solution as the inverse function x(y) is given by Page 87 of 90

Formation of differential equations Q 1 Form the differential equation of the following family of curves: a y= b y = asin(x+b) c y = Acosmx + Bsinmx, d y = c+cx+cx 2 e y = e 2X (a+bx) f e x (acosx +bsinx) g Y = ae 3X + be -2x h (x-a) 2 + (y-b) 2 = r 2 i Family of ellipses having foci on x-axis and centre at the origin Q 2 Family of circles touching x-axis at the origin Q3 Family of parabolas having vertex at the origin and axis along positive direction of x-axis Q4 Family of curves y = ae 2x + be -x Q5 Family of circles having centre on the x- axis and the radius is unity Q6 Family of circles in the first quadrant which touches the co-ordinate axis Variable Separable Q1 Solve Q 2 Solve Q3 Solve Q 4 Solve sin -1 = (x+y) Q 5 Solve cos -1 = (x+y) Q 6 log =3x+4y, given y =0 when x = 0 Page 88 of 90

Q7 Solve Q 8 Solve Q 4 Prove that is homogeneous and solve it Q 5 Solve (x+y)dy + (x-y)dx = 0 given y = 1 when x = 1 Q 6 Solve (x 2 -y 2 )dx+2xydy=0 given y = 1 when x = 1 Q 7 Solve (x 2 +xy)dy = (x 2 +y 2 )dx Q 8 Solve (3xy+y2)dx + (x 2 +xy)dy=0 Q 9 Solve (xdy-ydx) = dx Q 10 Solve (xdy-ydx)y =(ydx+xdy)x Q 11 Linear first order type Q 1 Solve Q 2 Solve Q 3 dx + xdy = Q 4 Solve Q 5 Solve xdy = Q 6 Solve Page 89 of 90

Q 7 Q 8 Q 9 Q 10 Q 11 Q 12 (1+x2)dy+2xydx=cotxdx Q 13 ; y = 0 when x = 1 Q 14 Q 15 given y = 2 when x = p/2 Q16 (tan-1y-x)dy=(1+y 2 )dx Q 17 Q18, given y = 0 when x = 0 Page 90 of 90