# Solve differential equation y^3y'+1/x y^4= sinx/x^4

Solve differential equation ${y}^{3}{y}^{\prime }+\frac{1}{x}{y}^{4}=\frac{\mathrm{sin}x}{{x}^{4}}$

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Let ${y}^{4}=z$ then, $4{y}^{3}{y}^{\prime }=\frac{dz}{dx}$
${y}^{3}{y}^{\prime }+\frac{1}{x}{y}^{4}=\frac{\mathrm{sin}x}{{x}^{4}}⇒\frac{1}{4}\frac{dz}{dx}+\frac{1}{x}z=\frac{\mathrm{sin}x}{{x}^{4}}$
$\frac{dz}{dx}+\frac{4}{x}z=\frac{4\mathrm{sin}x}{{x}^{4}}$
The given ordinary equation is of the form
$\frac{dz}{dx}+P\left(x\right)z=Q\left(x\right)$
Integrating factor: $IF={e}^{\int Pdx}={e}^{\int \frac{4}{x}}dx={x}^{4}$
$\frac{d}{dx}\left(IFz\right)=IFQ\left(x\right)$
$IF\ast z=\int IF\ast 1\left(x\right)dx$
${x}^{4}\ast z=\int \left({x}^{4}\right)\left(\frac{4\mathrm{sin}x}{x}\right)dx$
Multiply the original differential equation by the integral factor
$\frac{d}{dx}\left(IF\ast z\right)=IF\ast Q\left(x\right)$
$IF\ast z=\int IF\ast Q\left(x\right)dx$
${x}^{4}\ast z=\int \left({x}^{4}\right)\left(\frac{4\mathrm{sin}x}{{x}^{4}}\right)dx$
${x}^{4}\ast z=\int \left(4\mathrm{sin}x\right)dx$
${x}^{4}z=4\int \mathrm{sin}xdx$
${x}^{4}z=-4\mathrm{cos}x+c$
$z=\frac{-4\mathrm{cos}x}{{x}^{4}}+\frac{c}{{x}^{4}}$
Put the value of z in the obtained equation
$z=\frac{-4\mathrm{cos}x}{{x}^{4}}+\frac{c}{{x}^{4}}$
$z={y}^{4}$
${y}^{4}=\frac{-4\mathrm{cos}x}{{x}^{4}}+\frac{c}{{x}^{4}}$
$y=\sqrt[\frac{1}{4}]{\left(\frac{-4\mathrm{cos}x}{{x}^{4}}+\frac{c}{{x}^{4}}\right)}$