2326 Problem Sheet 1. Solutions 1. First Order Differential Equations. Question 1: Solve the following, given an explicit solution if possible:

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1 2326 Problem Sheet. Solutions First Order Differential Equations Question : Solve the following, given an explicit solution if possible:. 2 x = 4x3, () = 3, ( ) The given equation is a first order linear equation with integrating factor: ( ) 2 µ(x) = exp = exp( 2 ln x) = x 2 x Multipling the ODE b this factor we get: (x 2 ) = 4x x 2 = 2x 2 + C = 2x 4 + Cx 2 Imposing the initial condition () = 3 3 = 2 + C C = (x) = 2x 4 + x 2 2. t 2 2 = 0 This equation can be solved using separation of variables. The solution is (t) = 3/( t 3 + 3C), for some constant C. 3. x 2 = x ln We can rewrite the equation as follows: (x 2 + ) = x ln ln = x 2 + B separation of variables ln d = x x 2 + dx ln = 2 ln( + x2 ) + C The solution is left implicitl since we do not know an method to solve ln d. Esther Vergara Diaz, evd@maths.tcd.ie, see also

2 4. x = (ln x ln ). We have to make the substitution v = x, which implies v x + v = : x = (ln x ln ) = x ln x The solution is for some constant C (x) = xe, e C x 5. = ex +, (0) =. Using separation of variables, we get (x) = + 2e x + 2C 6. P = ap bp 2, a and b constants. Using separation of variables, and assuming the independent variable is x, we have: P ap bp 2 = ap bp 2 dp = We need to use partial fraction decomposition in the left hand side of the equation: ap bp 2 = A P + B a bp dx = A(a bp ) + BP P (a bp ) Equating coefficients, we get Aa = and B Ab = 0 then A = a and B = b a. Then ap bp 2 dp = a P dp + b a a bp dp = a ln P ln a bp a so P ln a bp = ax + C P a bp = Aeax, for some constant A. We can leave it in its implicit form. 2

3 7. = ( ) cos x, ( ) Using separation of variables, we get the following implicit result: + 2 = e x + C for some constant C. Imposing the initial condition 2 = + C C = Thus (x) + (x) 2 = e x + 8. x + 4x = 0, (2) = 7, ( ) Dividing both sides b x 2 and rewriting the equation a little, we get: ( ) 2 x = 4 x Let v(x) = x or vx =, recall that both v and are functions of x, we get can differentiate this equation to get v x + v = The ODE is rewritten as v(v x + v) = 4 v 2 We rewrite the equation so we can use separation of variables: vxv = 4 2v 2 xv = 4 + 2v2 v v 4 + 2v 2 dv = x dx Then: 4 ln(4 + 2v2 ) = ln x + C ln(4 + 2v 2 ) 4 = ln x + C (4 + 2v 2 ) D 4 = x So 4 + 2v 2 = A x 4 v2 = B x = B x 2 2x2 3

4 We do not take square roots since this will get two solutions. We leave the solution written implicitl. Now we impose the initial condition, (2) = 7: 49 = 2 (2) = B B = 228 Then, = ±( 228 x 2 2x2 ) 2 But since the initial condition was negative, we can ignore positive square root: 9. t + = t, ( ) = ( 228 x 2 2x2 ) 2 The given equation is a first order linear equation with integrating factor: ( µ(t) = exp Multipling the ODE we get: t x = x 2 ) = exp( ln(t + )) = t + t + ( ) = t t + + t = + + t = t ln( + t) + A (t) = t(t + ) (t + ) ln(t + ) + A(t + ) The solution is for some constant C. (x) = ( x + C)x,. ( + x 3 ) = 3x 2 + x 2 + x 5 We rewrite it as 3x2 + x 3 = x2 + x 5 + x 3 4

5 and observe that it is linear with P (x) = 3x2 + x 3 and Q(x) = x2 + x 5 + x 3. 3x 2 The integral factor is µ(x) = e +x 3 dx = e ln(+x3) = + x 3. Multipling the equation b µ(x), we get: ( ) + x 3 = x2 + x 5 ( + x 3 ) 2 = x2 + x 5 + 2x 3 + x 6 = 6x 2 + 6x x 3 + x 6 Integrating at both sides of the equation: + x 3 = 3 ln(x3 + ) + C (x) = x3 + (ln(x 3 + ) + D), 3 where D = 3C. Using separation of variables after writing the equation as = x2 cos x, we get the ln = or for some constant A 2. = 2 +, () = 2, ( ) We rewrite the equation as follows: x 2 cos xdx = x 2 sin x + 2 sin x 2x cos x + C (X) = Ae x2 sin x+2 sin x 2x cos x 2 + = and use separation of variables: 2 + d = dx = x + C 2 ln(2 + ) = x + C ln( 2 + ) = 2x + D 2 = Ee 2x = ± Ee 2x Imposing the initial condition we observe that the negative square root can be ignored and 4 = Ee 2 3 e 2 = E 5

6 So the solution is (x) = 3 e 2 e2x Question 2: (a) Show that the initial value problem t = 3, (0) =, has no solutions. Solving the equation using the method of separation of variables: (t) = Ct 3 Imposing the initial value we see that = (0) = C0 3 which is a contradiction, so this IVP does not have solution. (b) Show that the initial value problem t = 3, (0) = 0 has infinitel man solutions B separation of variables (t) = Ct 3, but all those curves go through zero, thus, (0) = C0 = 0, for ever C, so there are infinitel man solutions. (c) Show that the initial value problem t = 3, (3) = 5 has a unique solution. In this case (t) = Ct 3 5 = C3 3 C = 27 5, so the unique solution is (t) = 5 23 t3. Question 3: Classif the following ODEs (linear, homogenous, order).. s 2 x (s) + s 4 x(s) = s, first order linear nonhomogeneous ode. 2. t 2 (t) + (t) (t) =, second order nonlinear nonhomeneous ode. 3. x (x) + sin x(x) = 0, third order linear homogeneous ode. 4. x (x) + e x (x) = 0, second order linear homogeneous ode. 5. (t) (t) + e t2 (t) 2 = sin t, fourth order nonlinear nonhomegeneous ode. 6

7 Question 4: ( ) (a) Show that if = G(ax + b), then the change of variable w = ax + b reduce the equation to a separable one. If w = ax + b, then w = a + b or = w a. b We can rewrite the ODEs in terms of w: w a b variables: = G(w) w = G(w)b + a which can be solve b separation of G(w)b + a dw = dx = x + C (b) Consider the following change of variables w = 4x to solve the following differential equation: (4x + ) = 0 If w = 4x, then w 4 = w 4 w = 0 w = (w+5) w + 5 dw = dx = x+c ln(w + 5) = x + C w + 5 = De x w = De x 5 But w = 4x, then we can rewrite the solution in terms of : (x) = De x x ( ) To hand in on Februar, 3d. 7