Question

Find the particular integral solution of d³y/dx³-4dy/dx=x+3cosx+e^-2x

Answer 1

CORRECTED SOLUTION

$\frac{d^3y}{dx^3}-4\frac{dy}{dx}=x+3 \cos x+e^{-2x}$

We have a linear non-homogeneous ODE with constant coefficients in its homogeneous part. The characteristic values are the roots of the algebraic equation:

$\lambda^3-4\lambda=0$

They are $\lambda_1=0$ , $\lambda_2=2$ and $\lambda_3=-2$

Each of the terms in the right-hand side x, 3cos x and e^-2x are linear combinations of terms of the form $P(x)e^{\lambda x}$ with P(X) is a polynomial. We may seek the partial solutions of the same form $Q(x)e^{\lambda x}$ with degree of Q equal to degree of P, if the corresponding $\lambda$ is non-characteristic, or greater by 1 (up to the multiplicity of $\lambda$), if else. Furthermore, as the order of the equation is odd (3), we are to replace cosine to sine.

So, let us seek a solution of the form $y=a_1x^2+a_2 \sin x+a_3xe^{-2x}$. Then

$y'=2a_1x+a_2 \cos x-2a_3xe^{-2x}+a_3e^{-2x}$

$y''=2a_1-a_2 \sin x+4a_3xe^{-2x}-4a_3e^{-2x}$

$y'''=-a_2 \cos x-8a_3xe^{-2x}+12a_3e^{-2x}$

$y'''-4y'=-8a_1x-5a_2\cos x+8a_3e^{-2x}=x+3 \cos x+e^{-2x}$

Comparing the corresponding coefficients, we get:

a₁=-1/8, a₂=-3/5 and a₃=1/8.

Answer. $y=-\frac{1}{8}x^2-\frac{3}{5} \sin x+\frac{1}{8}xe^{-2x}$ .