Answer to Question #221980 in Differential Equations for Nikhil kumar

Question

d^2 / dx^2 - 2dy / dx= 3e^x sinx by method of undetermined coefficients

Answer 1

$d^2y / dx^2 - 2dy / dx= 3e^x\sin x$

Find the eigenvectors of the differential operator $d^2/dx^2 - 2d/dx$. Since this operator is linear and with constant coefficients, the eigenvectors may be found of the following form: $e^{\lambda x}$.

$(d^2/dx^2 - 2d/dx)e^{\lambda x}=(\lambda^2-2\lambda)e^{\lambda x}$

Therefore, the null-space o f this operator is generated by the functions $e^{\lambda x}$ with such $\lambda$, that $\lambda^2-2\lambda=0$, i.e. $\lambda_1=0$ and $\lambda_2=2$.

The general solution of linear ODE is the sum of a partial solution and an arbitrary function from the null space of the differential operator, that is, the function $ae^{2x}+b$ with any a,b.

Since $3e^x\sin x$ has neither the form $p_1(x)e^{\lambda_1x}$, nor the form $p_2(x)e^{\lambda_2x}$ with $p_1(x),\,p_2(x)$ polynomials on x, but has the form $e^{\lambda x}\sin (ax)$, we can seek a partial solution to be of the form

$y=e^x(a_1\sin x+a_2\cos x)$

Let's differentiate:

$y'=e^x(a_1\sin x+a_2\cos x+a_1\cos x-a_2\sin x)=e^x((a_1-a_2)\sin x+(a_1+a_2)\cos x)$

$y''=e^x(((a_1-a_2)-(a_1+a_2))\sin x+((a_1-a_2)+(a_1+a_2))\cos x)=$

$=e^x(-2a_2\sin x+2a_1\cos x)$

Hence

$y''-2y'=e^x(-2a_2\sin x+2a_1\cos x)-2e^x((a_1-a_2)\sin x+(a_1+a_2)\cos x)=$

$=e^x(-2a_1\sin x-2a_2\cos x)=3e^x\sin x$

Therefore, $a_1=-3/2$, $a_2=0$.

Answer. $y=-\frac{3}{2}e^x\sin x+ae^{2x}+b$