2xy^2+4=2(3-x^2y)y' y(-1)=8

Answered question

2022-01-19

2xy^2+4=2(3-x^2y)y' y(-1)=8

Answer & Explanation

nick1337

nick1337

Expert2022-02-01Added 777 answers

The given differential equation is:

2xy2+4=2(3x2y)y     y(1)=8

The above differential equation may be exprssed as:

(xy2+2dx+(x2y3)dy=0. whicj is of the form Mdx+Ndy=0

Here, we have

My=y(xy2+2)=2xy, and Nx=x(x2y3)=2xy

Since My=Nx, therefore the different equation is exact.

Hence, the required solution is:

y=constantMdx+ (terms of n containing x)dy=c

(xy2+2)dx+(3)dy=c

x2y22+2x3y=c

Given at x=1,y=8, therefore c=32224=6

Hence, the required solution of the differential equation is:

x2y22+2x3y=6

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