Answer to Question #316133 in Differential Equations for sandip

Question

Use method of undetermined coefficients method to solve the following

y''+y=5xsinx,y(0)=0,y'(0)=1

Answer 1

y''+y=5xsinx,y(0)=0,y'(0)=1

The solutions to a nonhomogeneous equation are of the form

$y(x) = y_c(x) + y_p(x)$ ,

where $y_c$ is the general solution to the associated homogeneous equation and $y_p$ is a particular solution.

The associated homogeneous equation:

$y''+y=0$

The general solution of this equation is determined by the roots of the characteristic equation:

$\lambda^2+1=0$

$\lambda=\pm i$ , where $i=\sqrt{-1}$ .

$y_c(x)=c_1\cos{x}+c_2\sin{x}$ .

The particular solution of the differential equation:

$y_p(x)=x((Ax+B)\sin{x}+(Cx+D)\cos{x})$

$y_p'=(2 A x + B - C x^2 - D x)\sin{x} +$$(A x^2 + B x + 2 C x + D)\cos{x}$

$y_p''= (-A x^2 + 2 A - B x - 4 C x - 2 D)\sin{x} +$$(4 A x + 2 B - C x^2 + 2 C - D x)\cos{x}$

Now put these into the original differential equation to get:

$(-A x^2 + 2 A - B x - 4 C x - 2 D)\sin{x} +$$(4 A x + 2 B - C x^2 + 2 C - D x)\cos{x} +$$x((Ax+B)\sin{x}+(Cx+D)\cos{x})=$$5x\sin{x}$

Equating coefficients, we get

$A=0$; $D=0$; $C=-\frac54$; $B=\frac54$.

$y_p(x)=\frac54x(\sin{x}-x\cos{x})$

$y=c_1\cos{x}+c_2\sin{x}+\frac54x(\sin{x}-x\cos{x})$

$y(0)=c_1=0$

$y’(0)=c_2=1$

Answer:

$y=\sin{x}+\frac54x(\sin{x}-x\cos{x})$ .