Question

a) Reduce the differential equation to separable form
and then solve
(x + 2y)dx + (2x + y)dy = 0, y(0) = 1​

Answer 1

$\underline{\textsf{Given:}}$

$\mathsf{(x+2y)dx+(2x+y)dy=0,\;\;y(0)=1}$

$\underline{\textsf{To find:}}$

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

$\underline{\textsf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{(x+2y)dx+(2x+y)dy=0}$

$\textsf{This can be written as}$

$\mathsf{x\,dx+2\,ydx+2\,xdy+y\,dy=0}$

$\mathsf{x\,dx+y\,dy=-2(x\,dy+y\,dx)}$

$\mathsf{x\,dx+y\,dy=-2\,d(xy)}$

$\mathsf{Integrating,}$

$\mathsf{\displaystyle\int\,x\,dx+\int\,y\,dy=-2\int\,d(xy)}$

$\mathsf{\dfrac{x^2}{2}+\dfrac{y^2}{2}=-2\,xy+C}$

$\mathsf{But\;y(0)=1}$

$\implies\mathsf{0+\dfrac{1}{2}=0+C}$

$\implies\mathsf{C=\dfrac{1}{2}}$

$\therefore\textsf{The solution is}$

$\mathsf{\dfrac{x^2}{2}+\dfrac{y^2}{2}=-2\,xy+\dfrac{1}{2}}$

$\mathsf{x^2+y^2=-4\,xy+1}$

$\boxed{\mathsf{x^2+y^2+4\,xy-1=0}}$

$\underline{\textsf{Find more:}}$

Solve x(x-y)dy+y^2dx=0​

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Solve dy/dx=e^3x-2y+ x^2 e^-2y

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