Answer to Question #175243 in Differential Equations for Three

Question

Using the method of variation of parameters find the general solution of the differential equation x^2 y'' - 3xy' + 3y = 12x^4 given that y1 = x and y2 = x^3 are solutions of the corresponding homogenous equation.

Answer 1

$x^{2}y^{''}-3xy^{'}+3y=12x^{4}$

$\Rightarrow y^{''}-3\frac{y^{'}}{x}+3\frac{y}{x^{2}}=12x^{2}$ $[ y^{''}+ay^{'}+by=R]$

Hence Wronskian (W)=$y_{1}y_{2}^{'}-y_{2}y_{1}^{'}=x 3x^{2}-x^{3}1=2x^{3} \neq 0$

Particular solution is $xf(x)+x^{3}g(x)$ where $f(x)=-\int\frac{y_{2} R}{W}dx$ $=-\int \frac{x^{3}12x^{2}}{2x^{3}} dx=-2x^{3}$

$g(x)=\int \frac{y_{1}R}{W} dx=\int\frac{x 12x^{2}}{x^{3}}dx=6x$

Therefore particular solution is $-2x^{3} x+x^{3}6x=4x^{4}$

The required solution is $y=c_{1}x+c_{2}x^{3}+4x^{4}$ where $c_{1},c_{2}$ are arbitrary constants.