Let y=y(x) be the solution of the differential equation (dy)/(dx)+5x(x^5+1)y=(x^5+1)^2x^7,…

Question

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}+\frac{5}{x({x}^{5}+1)}y=\frac{{({x}^{5}+1)}^{2}}{{x}^{7}},x>0$. If $y(1)=2$, then $y(2)$ is equal to :

Accepted Answer

I.F $={e}^{\int \frac{5dx}{x({x}^{5}+1)}}={e}^{\int \frac{5{x}^{-6}dx}{({x}^{-5}+1)}}$

Put, $1+{x}^{-5}=t\Rightarrow -5{x}^{-6}dx=dt$

$	herefore $ ${e}^{\int -\frac{dt}{t}}={e}^{-\ln t}=\frac{1}{t}=\frac{{x}^{5}}{1+{x}^{5}}$

$\begin{gathered} y\cdot \frac{{x}^{5}}{1+{x}^{5}} =\int \frac{{x}^{5}}{(1+{x}^{5})}	imes \frac{{(1+{x}^{5})}^{2}}{{x}^{7}}dx \ \ =\int {x}^{3}dx+\int {x}^{-2}dx \end{gathered}$

$y\cdot \frac{{x}^{5}}{1+{x}^{5}}=\frac{{x}^{4}}{4}-\frac{1}{x}+c$

$x=1\Rightarrow y=2$

$\begin{gathered} 2\cdot \frac{1}{2}=\frac{1}{4}-1+c \ c=\frac{7}{4} \ y\cdot \frac{{x}^{5}}{1+{x}^{5}}=\frac{{x}^{4}}{4}-\frac{1}{x}+\frac{7}{4} \ x=2 \ y\cdot (\frac{32}{33})=\frac{21}{4} \ y=\frac{693}{128} \end{gathered}$

Solutions · 1