Calculus
posted by Ella on .
Solve the IVP
{y'=(sinx)y + xexp(cosx)
y(0) = 1
Please show me step by step. Do I need to find out if this is exact? Do I use an integrating factor?

Solve the homogeneous equation first:
y_h' =  sin(x) y_h >
y_h(x) = K exp[cos(x)]
Then, to find the solution, you use the variation of the constant method, i.e. take the solution of the homogeneous solution y_h and there you replace K by an unknown function K(x):
y(x) = K(x) exp[cos(x)]
If you substitute this in the differential equation then what happens is that only the term proportional to K' survives. Due to the product rule the terms proportional to K are what you would get if K were a constant, but these terms will satisfy the homogeneous equation, so they will sum to zero.
So, what you get is:
K' Exp[cos(x)] = x Exp[cos(x)] >
K = 1/2 x^2 + c >
y(x) = (1/2 x^2 + c)Exp[cos(x)]