Solve differential equation1+y^2+xy^2)dx+(x^2y+y+2xy)dy=0

arenceabigns 2021-03-07 Answered

Solve differential equation \((1+y^2+xy^2)dx+(x^2y+y+2xy)dy=0\)

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Expert Answer

estenutC
Answered 2021-03-08 Author has 20744 answers

First Order Bernoulli differential equation is of the form \(y'+p(x)y=q(x)y^n\)
\((1+y^2+xy^2)+(x^2y+y+2xy)\ \frac{dy}{dx}=0\)
Put \(\frac{dy}{dx}= y'\)
\((1+y^2+xy^2)+(x^2y+y+2xy)\ y'=0\) (1)
Now converting the above equation (1) in the form
\(y'+p(x)y=q(x)y^n\)
Dividing the whole equation (1) by \((x^2y+y+2xy)\)
\((\frac{1+y^2+xy^2}{x^2y+y+2xy}+\frac{x^2y+y+2xy}{x^2y+y+2xy})y'=0\)
\((\frac{1+y^2(x+1)}{y(x^2+1+2x)})+y'=0\)
\(\frac{1+y^2(x+1)}{y(x+1)^2}+y'=0\)
\(\frac{1}{y(x+1)^2}+\frac{y^2(x+1)}{y(x+1)^2}+y'=0\)
\(\frac{1}{x+1}y+y'= -\frac{1}{x+1}^2 y^{-1}\)
\(y'+\frac{1}{x+1}y= - \frac{1}{x+1}^2 y^{-1}\)
\(p(x)= \frac{1}{x+1}\)
\(q(x)= -\frac{1}{x+1}^2\)
\(n= -1\) Now, substitute \(v=y^{1−n}\) in the Bernoulli equation, we get
\(v= y^{1-n}\)
\(y= v^{\frac{1}{1}-n}\)
\(\frac{1}{1-n}v'+p(x)v= q(x)\)
Solving by putting all the values we get
\(\frac{v'}{2}+\frac{v}{x+1}= −\frac{1}{x+1}^2\)
Put \(v= y^{2}\)
and then solve for v
\(v= -\frac{ 2x}{x^2+1+2x}+\frac{c_1}{x^2+1+2x}\)
Substituting back the value of y
\(y=\sqrt{-\frac{2x}{x^2+1+2x}+\frac{c_{1}}{x^2+1+2x}}\)
\(y=\sqrt{-2x+\frac{c_1}{x^2+2x+1}}\)

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