# The indicated functions are known linearly independent solutions of the associated homogeneous...

## Question:

The indicated functions are known linearly independent solutions of the associated homogeneous differential equation on {eq}\displaystyle (0,\ \infty) {/eq}. Find the general solution of the given non homogeneous equation.

{eq}\displaystyle x^2 y" + x y' + \bigg(x^2 - \frac{1}{4} \bigg) y = x^ {\frac{3}{2}},\ y_1 = x^{\frac{-1}{2}} \cos x,\ y_2 = x^{\frac{-1}{2}} \sin x {/eq}.

## Variation of Parameters:

The particular solution of a differential equation {eq}y''+p(x) y'+q(x)y=r(x) {/eq} is given by

{eq}y_p(x)=A(x)y_1(x)+B(x)y_2(x), {/eq}

where A(x) and B(x) are given by:

{eq}\displaystyle A(x)=-\int \frac{y_2(x) r(x)}{ W(y_1, \ y_2) } \ dx \\ \displaystyle B(x)=\int \frac{y_1(x) r(x)}{ W(y_1, \ y_2) } \ dx \\ {/eq}

and {eq}y_1(x), \ y_2(x) {/eq} are fundamental solutions to the associated homogeneous differential equation.

## Answer and Explanation:

We are given a non-homogeneous DE {eq}\displaystyle x^2 y'' + x y' + \bigg(x^2 - \frac{1}{4} \bigg) y = x^ {\frac{3}{2}}. {/eq}

The known linearly independent solutions of the associated homogeneous DE,

{eq}y_1 = x^{\frac{-1}{2}} \cos x, \ y_2 = x^{\frac{-1}{2}} \sin x {/eq}

The complimentay function is {eq}y_c(x)=C_1 x^{\frac{-1}{2}} \cos x + C_2 x^{\frac{-1}{2}} \sin x. {/eq}

Their wronskian of the fundamental solutions is

{eq}\begin{align} \displaystyle W(y_1, \ y_2) &=\begin{vmatrix} x^{\frac{-1}{2}} \cos x & x^{\frac{-1}{2}} \sin x \\ -x^{\frac{-1}{2}} \sin x-\frac{1}{2}x^{\frac{-3}{2}} \cos x & x^{\frac{-1}{2}} \cos x-\frac{1}{2}x^{\frac{-3}{2}} \sin x \end{vmatrix}\\ \displaystyle &=x^{\frac{-1}{2}} \cos x \left ( x^{\frac{-1}{2}} \cos x-\frac{1}{2}x^{\frac{-3}{2}}\sin x \right ) - x^{\frac{-1}{2}} \sin x \left ( -x^{\frac{-1}{2}} \sin x-\frac{1}{2}x^{\frac{-3}{2}}\cos x \right )\\ \displaystyle &=x^{-1} \cos^2 x-\frac{1}{2}x^{-2}\sin x \cos x+x^{-1} \sin^2 x+\frac{1}{2}x^{-2}\sin x \cos x\\ \displaystyle &=x^{-1} \left (\cos^2 x+\sin^2 x \right )\\ \displaystyle &=x^{-1} \end{align} {/eq}

The particular solution is

{eq}\begin{align} \displaystyle y_p(x) &=A(x)y_1(x)+B(x)y_2(x)\\ \displaystyle &=A(x) x^{\frac{-1}{2}} \cos x+B(x) x^{\frac{-1}{2}} \sin x, \end{align} {/eq}

where A(x) and B(x) are given by:

{eq}\displaystyle A(x)=-\int \frac{y_2(x) r(x)}{ W(y_1, \ y_2) } \ dx \\ \displaystyle B(x)=\int \frac{y_1(x) r(x)}{ W(y_1, \ y_2) } \ dx \\ {/eq}

{eq}\begin{align} \displaystyle A(x) &=-\int \frac{x^{\frac{-1}{2}} \sin x x^ {\frac{3}{2}} }{ x^{-1} } \ dx \\ \displaystyle &=-\int \frac{x \sin x }{ x^{-1} } \ dx \\ \displaystyle &=-\int x^2 \sin x \ dx \\ \displaystyle &=- \left [x^2 \int \sin x \ dx - \int \left ( \frac{d(x^2)}{dx} \int \sin x \ dx \right ) \ dx \right ] \\ \displaystyle &=- \left [-x^2 \cos x +2 \int \left ( x \cos x \right ) \ dx \right ] \\ \displaystyle &=x^2 \cos x -2 \int \left ( x \cos x \right ) \ dx \\ \displaystyle &=x^2 \cos x -2 \left ( x \int \cos x \ dx - \int \left ( \frac {dx}{dx} \int \cos x \ dx \right ) \ dx \right ) \\ \displaystyle &=x^2 \cos x -2 \left ( x \sin x - \int \sin x \ dx \right ) \\ \displaystyle &=x^2 \cos x -2 \left ( x \sin x + \cos x \right ) \\ \displaystyle &=x^2 \cos x -2 x \sin x -2 \cos x \\ \end{align} {/eq}

{eq}\begin{align} \displaystyle B(x) &=\int \frac{x^{\frac{-1}{2}} \cos x x^ {\frac{3}{2}} }{ x^{-1} } \ dx \\ \displaystyle &=\int x^2 \cos x \ dx \\ \displaystyle &=x^2 \int \cos x \ dx - \int \left ( \frac{d(x^2)}{dx} \int \cos x \ dx \right ) \ dx \\ \displaystyle &=x^2 \sin x -2 \int \left ( x \sin x \right ) \ dx \\ \displaystyle &=x^2 \sin x -2 \left ( x \int \sin x \ dx - \int \left ( \frac {dx}{dx} \int \sin x \ dx \right ) \ dx \right ) \\ \displaystyle &=x^2 \sin x -2 \left (- x \cos x + \int \cos x \ dx \right ) \\ \displaystyle &=x^2 \sin x -2 \left ( -x \cos x+ \sin x \right ) \\ \displaystyle &=x^2 \sin x +2 x \cos x -2 \sin x \\ \end{align} {/eq}

Hence, the particular solution is

{eq}\begin{align} \displaystyle y_p(x) &=(x^2 \cos x -2 x \sin x -2 \cos x ) x^{\frac{-1}{2}} \cos x + ( x^2 \sin x +2 x \cos x -2 \sin x ) x^{\frac{-1}{2}} \sin x\\ \displaystyle &=(x^{\frac{3}{2}} \cos^2 x -2 x^{\frac{1}{2}} \sin x \cos x -2 x^{\frac{-1}{2}} \cos^2 x ) + ( x^{\frac{3}{2}} \sin^2 x +2 x^{\frac{1}{2}} \sin x \cos x -2 x^{\frac{-1}{2}} \sin^2 x ) \\ \displaystyle &=x^{ \frac{3}{2} } - 2 x^{ \frac{-1}{2} } \\ \end{align} {/eq}

Therefore, the general solution is

{eq}\begin{align} \displaystyle y(x) &=y_c(x)+y_p(x)\\ \displaystyle &=\color{blue}{C_1 x^{\frac{-1}{2}} \cos x + C_2 x^{\frac{-1}{2}} \sin x+x^{ \frac{3}{2} } - 2 x^{ \frac{-1}{2} }} \end{align} {/eq}

#### Learn more about this topic:

from Calculus: Help and Review

Chapter 13 / Lesson 6