Math, asked by MuhammadHilal, 6 months ago

(y-2x^3)dx - x(1-xy)dy=0

Answers

Answered by MaheswariS
4

\underline{\textbf{Given:}}

\mathsf{(y-2x^3)dx-x(1-xy)dy=0}

\underline{\textbf{To find:}}

\textsf{Solution of the differential equation}

\underline{\textbf{Solution:}}

\textsf{We apply Variable separable method to solve}

\textsf{the differential equation}

\mathsf{Consider,}

\mathsf{(y-2x^3)dx-x(1-xy)dy=0}

\textsf{This can be written as}

\mathsf{(y\,dx-2x^3\,dx-x\,dy+x^2ydy=0}

\mathsf{(y\,dx-x\,dy=2x^3\,dx-x^2ydy}

\mathsf{y\,dx-x\,dy=x^2(2x\,dx-ydy)}

\mathsf{\dfrac{y\,dx-x\,dy}{x^2}=2x\,dx-ydy}

\mathsf{\dfrac{x\,dy-y\,dx}{x^2}=ydy-2x\,dx}

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

\mathsf{dt=ydy-2x\,dx}

\mathsf{Integrating,}

\mathsf{\displaystyle\int\,dt=\int\,y\,dy-2\int\,x\,dx}

\mathsf{t=\dfrac{y^2}{2}-2\left(\dfrac{x^2}{2}\right)+C}

\implies\boxed{\mathsf{\dfrac{y}{x}=\dfrac{y^2}{2}-x^2+C}}

\textsf{which is the required solution}

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