A nonhomogeneous second-order linear equation and a complementary function $y}_{c$ are given. Find a particular solution of the equation.

$x}^{2}y{}^{\u2033}-4x{y}^{\prime}+6y={x}^{3};\text{}{y}_{c}={c}_{1}{x}^{2}+{c}_{2}{x}^{3$

Kendall Daniels
2022-03-28
Answered

A nonhomogeneous second-order linear equation and a complementary function $y}_{c$ are given. Find a particular solution of the equation.

$x}^{2}y{}^{\u2033}-4x{y}^{\prime}+6y={x}^{3};\text{}{y}_{c}={c}_{1}{x}^{2}+{c}_{2}{x}^{3$

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Gia Edwards

Answered 2022-03-29
Author has **12** answers

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,

$W\left(x\right)=[({x}^{2},{x}^{3}),(\frac{d}{dx}\left({x}^{2}\right),\frac{d}{dx}\left({x}^{3}\right)]=\left[\begin{array}{cc}{x}^{2}& {x}^{3}\\ 2x& 3{x}^{2}\end{array}\right]={x}^{4}$

Dividing the differential equation by the leading terms coefficient

$x}^{2}y{}^{\u2033}-4x{y}^{\prime}+6y={x}^{3$

$\frac{{x}^{2}y{}^{\u2033}-4x{y}^{\prime}+6y}{{x}^{2}}=\frac{{x}^{3}}{{x}^{2}}$

$y{}^{\u2033}-\frac{4{y}^{\prime}}{x}+\frac{6y}{{x}^{2}}=x$

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,

${y}_{p}\left(x\right)={v}_{1}\left(x\right){y}_{1}\left(x\right)+{v}_{2}\left(x\right){y}_{2}\left(x\right)$

${v}_{1}\left(x\right)=-\int \frac{x\left({x}^{3}\right)}{{x}^{4}}dx=-\int 1dx=-x$

${v}_{2}\left(x\right)=\int \frac{x\left({x}^{2}\right)}{{x}^{4}}dx=-\int \frac{1}{x}dx=\mathrm{ln}x$

${y}_{p}\left(x\right)=x\left({x}^{2}\right)+\left(\mathrm{ln}x\right)\left({x}^{3}\right)$

${y}_{p}\left(x\right)={x}^{3}(1+\mathrm{ln}x)$

It is known that,

Now, computing the Wro

ian,

Dividing the differential equation by the leading terms coefficient

Then,

The particular solution is given by,

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Hence:

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That being said, solving the equation I try as follows:

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$x{y}^{\prime}+2y\mathrm{log}y-4{x}^{2}y=0;$

$y(1)=1$

I want to use the substitution $v=\mathrm{log}y$. Which implies that

${v}^{\prime}=\frac{dv}{dy}\frac{dy}{dx}=\frac{1}{y}{y}^{\prime}$

Hence:

${y}^{\prime}=y{v}^{\prime}$

That being said, solving the equation I try as follows:

$xy{v}^{\prime}+2yv-4{x}^{2}y=0$

$x{v}^{\prime}+2v=4{x}^{2}$

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How can I use seperation of variables here? Or how can I solve this thing ?

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