Use any appropriate method in finding the indicated solution of the following differential
equations.
[2π₯π¦ cos(π₯2) β 2π₯π¦ + 1]ππ₯ + [π ππ(π₯2) β π₯2]ππ¦ = 0
[2xycosβ‘x2β2xy+1]dx+[sinβ‘x2βx2]dy=0ββx[ysinβ‘x2βx2y+x]dx+ββy[ysinβ‘x2βx2y+x]dy=0d[ysinβ‘x2βx2y+x]=0ysinβ‘x2βx2y+x=Cy=Cβxsinβ‘x2βx2\left[ 2xy\cos x^2-2xy+1 \right] dx+\left[ \sin x^2-x^2 \right] dy=0\\\frac{\partial}{\partial x}\left[ y\sin x^2-x^2y+x \right] dx+\frac{\partial}{\partial y}\left[ y\sin x^2-x^2y+x \right] dy=0\\d\left[ y\sin x^2-x^2y+x \right] =0\\y\sin x^2-x^2y+x=C\\y=\frac{C-x}{\sin x^2-x^2}[2xycosx2β2xy+1]dx+[sinx2βx2]dy=0βxββ[ysinx2βx2y+x]dx+βyββ[ysinx2βx2y+x]dy=0d[ysinx2βx2y+x]=0ysinx2βx2y+x=Cy=sinx2βx2Cβxβ
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