# A nonhomogeneous second-order linear equation and a complementary

A nonhomogeneous second-order linear equation and a complementary function ${y}_{c}$ are given. Find a particular solution of the equation.
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Gia Edwards
Calculation:
It is known that,
${y}_{1}\left(x\right)={x}^{2}$ and ${y}_{2}\left(x\right)={x}^{3}$
Now, computing the Wro
ian,
$W\left(x\right)=\left[\left({x}^{2},{x}^{3}\right),\left(\frac{d}{dx}\left({x}^{2}\right),\frac{d}{dx}\left({x}^{3}\right)\right]=\left[\begin{array}{cc}{x}^{2}& {x}^{3}\\ 2x& 3{x}^{2}\end{array}\right]={x}^{4}$
Dividing the differential equation by the leading terms coefficient
${x}^{2}y{}^{″}-4x{y}^{\prime }+6y={x}^{3}$
$\frac{{x}^{2}y{}^{″}-4x{y}^{\prime }+6y}{{x}^{2}}=\frac{{x}^{3}}{{x}^{2}}$
$y{}^{″}-\frac{4{y}^{\prime }}{x}+\frac{6y}{{x}^{2}}=x$
Then,
$f\left(x\right)=x$
${v}_{1}\left(x\right)=-\int \frac{x{y}_{2}\left(x\right)}{W\left(x\right)}dx$ and ${v}_{2}\left(x\right)=\int \frac{x{y}_{2}\left(x\right)}{W\left(x\right)}dx$
The particular solution is given by,
${y}_{p}\left(x\right)={v}_{1}\left(x\right){y}_{1}\left(x\right)+{v}_{2}\left(x\right){y}_{2}\left(x\right)$
${v}_{1}\left(x\right)=-\int \frac{x\left({x}^{3}\right)}{{x}^{4}}dx=-\int 1dx=-x$
${v}_{2}\left(x\right)=\int \frac{x\left({x}^{2}\right)}{{x}^{4}}dx=-\int \frac{1}{x}dx=\mathrm{ln}x$
${y}_{p}\left(x\right)=x\left({x}^{2}\right)+\left(\mathrm{ln}x\right)\left({x}^{3}\right)$
${y}_{p}\left(x\right)={x}^{3}\left(1+\mathrm{ln}x\right)$