5. Suppose we are given the non-linear differential equation y' + r(x)y = x* . y7. To solve the equation: A) First, we would make the substitution v = yl i = yi B) Applying the substitution, we would get the following corresponding linear equation in v: u' + f ( x )v = x4 = 0'+ : C) We then find the solution of the differential equation in v, let's say the solution is v = Q(x) + c. Use the given solution for v to determine the solution of the original differential equation. v = Q (x) + c= yi = Q(x)+c=y= (Q(x)+ c)+ 6. Solve: (4t3x2 - 6t2x - 2t - 3)dt + (2t*x - 2t3)dx = 0,x(1) = 3 The equation appears to exact, with M = 4t3x2 - 6t2x - 2t - 3 and N = 2t*x - 2t3. Since My = 8t3x - 6t2 and N. = 8t3x - 6t2, the equation is exact. Then F(t, x) = [ Mat = S (4t3x2 - 6t2x - 2t - 3)dt = t*x2 - 2t3x - t2 - 3t + g(x) Then Fy = 2t*x - 2t3 + g'(x) = N, so 2t4x - 2t3 + g'(x) = (2t*x -2+3) Then g'(x) = 0, so g(x) = c1 The general solution then is F(t, x) = C2, so tax? - 2t3x - t2 - 3t + ci = Cz, or t*x2 - 2t3x - t2 - 3t = c Applying the initial condition: x(1) =3=9-6-1-3=cc=-1 So the specific solution is 14x2 - 2+3x - t2 - 3t= -1 Page 4 of 7Page 4 Test Exercise 5. Suppose we are given the non-linear differential equation y' + f(x)y = x2 . vy. To solve the equation: A) First, we would make the substitution v = B) Applying the substitution, we get the following corresponding linear equation in v: C) We then find the solution of the differential equation in v, let's say the solution is v = G(x) + c. Use the given solution for v to determine the solution of the original differential equation. Test Exercise 6. Solve: (6x cosy + ex - 9)dx + (-3x2 siny + xex + 4)dy = 0, y(1) = 0