Question

Solution for dy/dx+xsin 2 y=x^3 cos^2y

Answer 1

Rearrange the ODE as


`(\mathrm dy)/(\mathrm dx)+x\sin2y=x^3\cos^2y`

`\sec^2y(\mathrm dy)/(\mathrm dx)+x\sin2y\sec^2y=x^3`

Take
`u=\tan y`, so that
`(\mathrm du)/(\mathrm dx)=\sec^2y(\mathrm dy)/(\mathrm dx)`.

Supposing that
`|y|<\frac\pi2`, we have
`\tan^(-1)u=y`, from which it follows that


`\sin2y=2\sin y\cos y=2\frac u{√(u^2+1)}\frac1{√(u^2+1)}=(2u)/(u^2+1)`

`\sec^2y=1+\tan^2y=1+u^2`

So we can write the ODE as


`(\mathrm du)/(\mathrm dx)+2xu=x^3`

which is linear in
`u`. Multiplying both sides by
`e^(x^2)`, we have


`e^(x^2)(\mathrm du)/(\mathrm dx)+2xe^(x^2)u=x^3e^(x^2)`

`(\mathrm d)/(\mathrm dx)\bigg[e^(x^2)u\bigg]=x^3e^(x^2)`

Integrate both sides with respect to
`x`:


`\displaystyle\int(\mathrm d)/(\mathrm dx)\bigg[e^(x^2)u\bigg]\,\mathrm dx=\int x^3e^(x^2)\,\mathrm dx`

`e^(x^2)u=\displaystyle\int x^3e^(x^2)\,\mathrm dx`

Substitute
`t=x^2`, so that
`\mathrm dt=2x\,\mathrm dx`. Then


`\displaystyle\int x^3e^(x^2)\,\mathrm dx=\frac12\int 2xx^2e^(x^2)\,\mathrm dx=\frac12\int te^t\,\mathrm dt`

Integrate the right hand side by parts using


`f=t\implies\mathrm df=\mathrm dt`

`\mathrm dg=e^t\,\mathrm dt\implies g=e^t`

`\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)`

You should end up with


`e^(x^2)u=\frac12e^(x^2)(x^2-1)+C`

`u=\frac{x^2-1}2+Ce^(-x^2)`

`\tan y=\frac{x^2-1}2+Ce^(-x^2)`

and provided that we restrict
`|y|<\frac\pi2`, we can write


`y=\tan^(-1)\left(\frac{x^2-1}2+Ce^(-x^2)\right)`