Lesson 3: Linear differential equations of the first order Solve each of the following differential equations by two methods.

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1 Lesson 3: Linear differential equations of the first der Solve each of the following differential equations by two methods. Exercise 3.1. Solution. Method 1. It is clear that y + y = 3 e dx = e x is an integrating fact f this differential equation. Therefe, e x [y + y] = 3e x, [ye x ] = 3e x. The solution is therefe the following: ye x = c + 3 e x dx, simply, y = ce x + 3. Method 2. Consider a homogeneous equation The solution of this equation is y + y =. e x. To solve the initial equation, introduce a new variable v, to find the solution in the fm y = e x v. ( ) Then Substitution it f (*) yields Exercise 3.2. Solution. Method 1. It is clear that y = v e x e x v = e x [v v]. y + y = v e x = 3, v = 3e x, v = 3e x + c y = 3 + ce x. y 2y = x e ( 2)dx = e 2x is an integrating fact f this differential equation. Therefe, e 2x [y 2y] = xe 2x, [ye 2x ] = xe 2x. 1

2 2 The solution is therefe the following: ye 2x = c + xe 2x dx = c.5xe 2x.25e 2x simply, y = ce 2x.5x.25 Method 2. Consider a homogeneous equation The solution of this equation is y 2y =. e 2x. To solve the initial equation, introduce a new variable v, to find the solution in the fm y = e 2x v. Then Therefe, Exercise 3.3. y = v e 2x + 2e 2x v = e 2x [v + 2v]. y 2y = v e 2x = x, v = xe 2x, v = c.5xe 2x.25e 2x y = e 2x [c.5xe 2x.25e 2x ] = ce 2x.5x.25 xy + y = 2x + e x. Solution. Method 1. It is not difficult to see that the differential equation is exact. Therefe, rewriting it as xdy + ( y 2x e x) dx =, we obtain the solution x ( 2u e u) du + which in turn reduces to y xy x 2 e x + 1 = c 1, xy x 2 e x = c. xdv = c 1 Method 2. Assuming that x let us divide the both sides of differential equation to x. We obtain y + y x = 2 + ex x. Let us now consider the homogeneous differential equation It is clear that y + y x =. y = 1 x

3 3 is a solution. Introduce a new variable v, representing Exercise 3.4. y = v x. y = xv v x 2 v x = 2 + ex x, v = 2x + e x, v = x 2 + e x + c. xy = c + x 2 + e x. x 2 y xy = x Solution. Notice that x. one can divide the left right sides of differential equation to x 2. We have y y x = x 2. Let us solve the last differential equation by two methods. Method 1. Notice that, is an integrating fact. Therefe, 1 ( y y ) x x By integrating we obtain Method 2. The homogeneous equation has the solution Introduce a new variable v, representing e x 1 dx = e log x = 1 x = 1 x (1 + 4 x 2 ), ( y ) ( 1 = x x + 4 ) x 3 ( y = log x x) 2 x 2 + c, yx = x 2 log x 2 + cx 2. y y x =. y = x. y = xv. y = v + xv

4 4 By integrating, Exercise 3.5. Solution. xv = x 2, v = 1 x + 4 x 3. v = log x 2 x 2 + c. vx 2 = yx = x 2 log x 2 + cx 2. y + ay = b (a, b constants). Method 1. Notice that, e adx = e ax is an integrating fact. Therefe, e ax [y + ay] = be ax is an exact differential equation. ( e ax y ) = be ax, if a, if a e ax y = b a eax + c. y = b a + ce ax, otherwise y = bx + c. Method 2. The homogeneous equation (a ) has the solution Introducing new variable v put Therefe, Exercise 3.6. y + ay =. y = e ax. y = e ax v. y = ave ax + v e ax. v e ax = b, v = be ax v = b a eax + c. y = b a + ce ax. y + 1 x y = ex2.

5 is an integrating fact. Therefe, (e x p) = 2e x, 5 Solution. Method 1. Notice that, e x 1 dx = e log x = x is an integrating fact. Therefe, x (y + 1 ) x y = xe x2 is an exact differential equation. By integrating we obtain Method 2. The homogeneous equation has the solution Introducing new variable v put Exercise 3.7. Solution. Denoting y = p we obtain (xy) = xe x2. 2xy = e x2 + c. y + 1 x y = y = 1 x. y = v x. y = xv v x 2 v x = 2 ex, 2v = e x2 + c. 2xy = e x2 + c. y + y = 2. p x + p = 2. Let us now solve this linear differential equation by two methods. Method 1. Notice that e dx = e x by integrating p = 2 + c 1 e x. By integrating once again we obtain y = c 2 + 2x c 1 e x.

6 6 Method 2. The homogeneous equation p x + p = has the solution e x. Therefe, introducing a new variable v we find the solution as p = e x v. The solution is Integrating once again we have Exercise 3.8. Solution. Denoting y = p, we have p = e x v e x v. e x v = 2, v = c 1 + 2e x. p = c 1 e x + 2. y = c 2 + 2x c 1 e x. xy + y = x 2. xp x + p = x 2. Method 1. Notice that differential equation is exact. Therefe Rewriting it x integrating, we obtain ( u + 2)du + y = c 1 2 p xdv = 2x x2 2 + xp =.5c 1 4x x 2 + 2xp = c 1 p = c 1 2x + x2 4x 2x log x + x2 4 2x +.25c 2. 4y = 2c 1 log x + x 2 2x + c 2. Method 2. In the case x = the solution is trivial y = 2x + c. Assuming that x let us divide to x. We have Consider the homogeneous equation p x + p x = 1 2 x. p x + p x =. Its solution is p = x 1. Therefe, the solution of general (nonhomogeneous equation) is p = x 1 v. Then p = v x v x 2.

7 7 That is By integrating we have v x = 1 2 x. v = x 2 v =.25c 1 + x2 2 2x, p = c 1 4x + x 2 2 Integrating once again, we obtain y = c 1 4 log x + x2 4 2x +.25c 2 finally 4y = c 1 log x + x 2 2x + c 2. Exercise 3.9. y 2 dx + (y 2 x + 2xy 1)dy =. Solution. F convenience, rewrite this differential equation as follows: y 2 dx dy + (y2 x + 2xy 1) =. Taking into account that y, let us divide the equation to y 2. We have dx (1 dy + x + 2 ) = 1 y y 2. Solve this equation by the two methods. Method 1. Notice that e (1+2y 1 )dy = y 2 e y is an integrating fact. Therefe y 2 e y dx ( dy + x y 2 e y + 2ye y) = e y. By immediate integration, we obtain x y 2 e y du y xy 2 e y = c + e y e v dv = c 1, x = 1 y 2 (1 + ce y). Method 2. The homogeneous differential equation dx (1 dy + x + 2 ) = y has the solution x = e y y 2. Therefe, the solution of nonhomogeneous differential equation is x = e y 1 y 2 v,

8 8 x = ([ e y 1 y 2 2 y e y] v + v e y 1 y 2 ). v e y 1 y 2 = 1 y 2, v = e y, v = c + e y x = c 1 y 2 e y + 1 y 2. Exercise 3.1. y + (x y 3 2)y =. Solution. Rewrite the equation as follows yx + x y 3 2 =. Method 1. Notice that the differential equation is exact. Therefe, Method 2. follows x du + y (x v 3 2)dv = xy y4 4 2y =.25c 4xy y 4 8y = c. Assuming that y let us represent the differential equation as Consider first the homogeneous equation x + 1 y x = y2 2 y. x + 1 y x =. Its solution is x = y 1. Therefe, the general solution is x = v y. Therefe By integrating we have v = y4 4 x = v y v y 2. v y = y2 2 y, v = y y +.25c 4xy = y c. Exercise y y = 1, y() =.

9 9 Solution. F the homogeneous differential equation y y = we have the solution y = e x. Therefe, f nonhomogeneous equation we have finally y = e x v, y = ve x + e x v. e x v = 1, v = e x, v = c e x, y = ce x 1. c = (y + 1)e x. Substituting x = y =, we obtain c = 1. Thus, y = e x 1 is the solution of the problem. Exercise xy + y = 2x, y(1) = 1. Solution. Notice that the differential equation is exact. Therefe x ( 2u)du + y xdv = xy x 2 = c. Putting x = 1 y = 1 we find c =. Thus, finally Exercise xy x 2 = y = x. e x y + 2e x y = e x, y() = e. Solution. Multiplying the left right sides to e x we obtain y + 2e 2x y = e 2x. This is a linear differential equation. The homogeneous differential equation y + 2e 2x y = has the solution y = exp( exp(2x)). Therefe, the solution of the nonhomogeneous differential equation has solution Therefe y = e e2x v. y = 2ve e2x e 2x + e e2x v. e e2x v = e 2x, v = e 2x e e2x = e 2x+e2x.

10 1 By integrating we have v = c + e 2x+e2x dx = c e +e2x de 2x = c ee2x. y = ce e2x Putting x =, y =.5 + e 1 we obtain c = 1. Therefe, Exercise y = e e2x ( π ) (sin x)y + (cos x)y = cos 2x, y = Solution. If x = the solution of the differential equation is trivial as y = 1. Let us assume that x dividing both sides of the differential equation to sin x. Then we obtain y cos 2x + (cot x)y = sin x. The homogeneous differential equation y + (cot x)y = has a solution (sin x) 1. Therefe the solution of the nonhomogeneous differential equation is represented y = v sin x, y = (sin x)v (cos x)v sin 2. x v cos 2x = sin x sin x, v = cos 2x. By integrating we obtain 2v = c + sin 2x, 2y sin x = c + sin 2x, 2y = c sin x + 2 cos x Now, putting y =.5 x =.5π we obtain c = 1 the final solution is 2y = cos x. sin x address: vyachesl@inter.net.il