Question #320862

Use any appropriate method in finding the indicated solution of the following differential 

equations. 

[2π‘₯𝑦 cos(π‘₯2) βˆ’ 2π‘₯𝑦 + 1]𝑑π‘₯ + [𝑠𝑖𝑛(π‘₯2) βˆ’ π‘₯2]𝑑𝑦 = 0


1
Expert's answer
2022-03-31T05:39:40-0400

[2xycos⁑x2βˆ’2xy+1]dx+[sin⁑x2βˆ’x2]dy=0βˆ‚βˆ‚x[ysin⁑x2βˆ’x2y+x]dx+βˆ‚βˆ‚y[ysin⁑x2βˆ’x2y+x]dy=0d[ysin⁑x2βˆ’x2y+x]=0ysin⁑x2βˆ’x2y+x=Cy=Cβˆ’xsin⁑x2βˆ’x2\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}


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