Answer to Question #128879 in Differential Equations for Nikhil

Question

Given that sinx is a solution of the differential equation: L[y]=d^4y/dx^4+2d^3y/dx^3+6d^2ydx^2+2dy/dx+5y=0 and the 4 parameter family of solutions of it. Hence solve L[y]=x^2

Answer 1

$$Solution$$

To evaluate the particular solution $l[y]=x^2$ .

It was given that $\sin(x)$ is a solution of

$$\frac{d^4y}{dx^4}+2\frac{d^3y}{dx^3}+6\frac{d^2y}{dx^2}+2\frac{dy}{dx}+5y=0$$

Next,

$$\frac{d^4y}{dx^4}+2\frac{d^3y}{dx^3}+6\frac{d^2y}{dx^2}+2\frac{dy}{dx}+5y=x^2$$

For D.E $\frac{d^4y}{dx^4}+2\frac{d^3y}{dx^3}+6\frac{d^2y}{dx^2}+2\frac{dy}{dx}+5y=x^2$ , you need to try the solution

$y_{p_1}=Ax^2+Bx+C$ (Since $l[y]=x^2$ is of degree 2), Such that

$$\frac{d^2y}{dx^2}=A, \frac{dy}{dx}=B$$

$$\frac{d^3y}{dx^3}=\frac{d^4y}{dx^4}=0$$

$6A+2B+5(Ax^2+Bx+c)=x^2$

$5Ax^2+6A+5Bx+2B+5C=x^2$

Equating the coefficients of like powers yields:

$$5A=1 \implies A=\frac{1}{5}$$

$$5B=0 \implies B=0$$

$$6A+2B+5C=0 \implies 6A+5C=0 \implies 5C=-\frac{6}{5} \implies C=-\frac{6}{25}$$

Hence, evaluating $y_{p_1}$ yields

$$y_{p_1}=\frac{1}{5}x^2-\frac{6}{25}$$