Question

xdy=(y+4x^5 e^x^4)dx

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Answer 1

✯ $\textbf{Given:}$ ✯

$\mathsf{x\,dy=(y+4x^5\,e^{x^4})dx}$

✯ $\textbf{To find:}$ ✯

$\textsf{solution of the given differential equation}$

✯ $\textbf{Solution:}$ ✯

$\textbf{We apply variale separable method to solve the differential equation}$

✯ $\mathsf{Consider,}$ ✯

$\mathsf{x\,dy=(y+4x^5\,e^{x^4})dx}$

$\mathsf{x\,dy=y\,dx+4x^5\,e^{x^4}\,dx}$

$\mathsf{x\,dy-y\,dx=(x^2)4x^3\,e^{x^4}\,dx}$

$\mathsf{\dfrac{x\,dy-y\,dx}{x^2}=4x^3\,e^{x^4}\,dx}$

$\boxed{\begin{minipage}{4cm} \\\mathsf{Take\;t=x^4}\\\\\mathsf{\dfrac{dt}{dx}=4x^3}\\\\\mathsf{dt=4x^3\,dx}\\ \end{minipage}}$

$\boxed{\begin{minipage}{4cm} \\\mathsf{Take\;u=\dfrac{y}{x}}\\\\\mathsf{\dfrac{du}{dx}=\dfrac{x\frac{dy}{dx}-y.1}{x^2}}\\\\\mathsf{du=\dfrac{x\,dy-y\,dx}{x^2}}\\ \end{minipage}}$

$\implies\mathsf{du=e^t\,dt}$

$\mathsf{Integrating,}$

$\implies\mathsf{\int\,du=\int\,e^t\,dt}$

$\implies\mathsf{u=e^t+c}$

$\implies\boxed{\mathsf{\dfrac{y}{x}=e^{x^4}+c}}$✰

$\textsf{Which is the required solution}$✡

$\pink{\sf{\star\;Note}}$

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