 Helen Lewis

2021-12-29

${y}^{\prime }=x-2y\mathrm{cot}2x$ Piosellisf

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Step 1
Given that the equation ${y}^{\prime }=x-2y\mathrm{cot}2x$
${y}^{\prime }+\left(2\mathrm{cot}2x\right)y=x$
This is a linear differential equation in y.
$\frac{dy}{dx}+Py=Q$
Here, $P=2\mathrm{cot}2x,Q=x$
$\int Pdx=\int 2\mathrm{cot}2xdx=2\left[\frac{\mathrm{log}\mathrm{sin}2x}{2}\right]=\mathrm{log}\mathrm{sin}2x$
Thus, integral factor,
Step 2
Hence the solution is

$y\left(\mathrm{sin}2x\right)=\int x\left(\mathrm{sin}2x\right)dx+C$
$y\left(\mathrm{sin}2x\right)=\left[x\left(\frac{-\mathrm{cos}2x}{2}\right)+\left(1\right)\left(\frac{-\mathrm{sin}2x}{4}\right)\right]+C$
$y\left(\mathrm{sin}2x\right)=\left[\frac{-x\mathrm{cos}2x}{2}-\frac{\mathrm{sin}2x}{4}\right]+C$
$y\mathrm{sin}2x=\frac{1}{4}\left(-2x\mathrm{cos}2x-\mathrm{sin}2x\right)+C$ is the required solution. ol3i4c5s4hr

Expert

${y}^{\prime }=x-2y\mathrm{cot}\left(2x\right)$
${y}^{\prime }+2y\mathrm{cot}\left(2x\right)=x$
By inspection, the equation is a linear DE in the form of:
${y}^{\prime }+yP\left(x\right)=Q\left(x\right)$
where:
The integrating factor, i.f. is:
$i.f.={e}^{\int P\left(x\right)dx}$
$i.f={e}^{\int 2\mathrm{cot}\left(2x\right)dx}$
Note:
$i.f.={e}^{\mathrm{ln}\left(\mathrm{sin}\left(2x\right)\right)}$
Note: ${e}^{\mathrm{ln}u}=u$
$i.f=\mathrm{sin}\left(2x\right)$
Substituting the i.f., we get:
$y\mathrm{sin}\left(2x\right)=\int ×\mathrm{sin}\left(2x\right)dx+C$
For $\int ×\mathrm{sin}\left(2x\right)dx$, use integration by parts.
In integration by parts, choose u in this order: LIATE
Logarithmic
Inverse
Algebraic
Trigonometric
Exponential
$\int ×\mathrm{sin}\left(2x\right)dx$
Using integration by parts:
Let: $u=x$
$du=dx$
$dv=\mathrm{sin}\left(2x\right)dx$
$v=-\frac{1}{2}\mathrm{cos}\left(2x\right)$
$\int udv=uv-\int vdu$
$\int ×\mathrm{sin}\left(2x\right)dx=-\frac{1}{2}×\mathrm{cos}\left(2x\right)-\int -\frac{1}{2}\mathrm{cos}\left(2x\right)dx$
$\int ×\mathrm{sin}\left(2x\right)dx=-\frac{1}{2}×\mathrm{cos}\left(2x\right)+\frac{1}{4}\mathrm{sin}\left(2x\right)$
Substituting the value of $\int ×\mathrm{sin}\left(2x\right)dx$, we get: Vasquez

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