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

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

Answers (1)

2021-03-08
First Order Bernoulli differential equation is of the form \(y'+p(x)y=q(x)y^n\)
\((1+y^2+xy^2)+(x^2y+y+2xy)dy/dx=0\)
Put \(dy/dx= y'\)
\((1+y^2+xy^2)+(x^2y+y+2xy)y'=0\) (1)
Now converting the above equation (1) in the form
\(y'+p(x)y=q(x)y^n\)
Dividing the whole equation (1) by \((x^2y+y+2xy)\)
\(((1+y^2+xy^2))/(x^2y+y+2xy)+((x^2y+y+2xy))/(x^2y+y+2xy)y'=0\)
\(((1+y^2(x+1)))/((y(x^2+1+2x)))+y'=0\)
\((1+y^2(x+1))/(y(x+1)^2)+y'=0\)
\(1/(y(x+1)^2)+(y^2(x+1))/(y(x+1)^2)+y'=0\)
\(1/(x+1)y+y'= -1/(x+1)^2 y^-1\)
\(y'+1/(x+1)y= - 1/(x+1)^2 y^-1\)
\(p(x)= 1/(x+1)\)
\(q(x)= -1/(x+1)^2\)
n= -1 Now, substitute \(v=y^(1−n)\) in the Bernoulli equation, we get
\(v= y^(1-n)\)
\(y= v^(1/(1-n))\)
\(1/(1-n)v'+p(x)v= q(x)\)
Solving by putting all the values we get
\((v')/2+v/(x+1)= −1/(x+1)^2\)
Put \(v= y^2
and then solve for v
\(v= - 2x/(x^2+1+2x)+c_1/(x^2+1+2x)\)
Substituting back the value of y
\(y= sqrt(-(2x)/(x^2+1+2x)+c_1/(x^2+1+2x))\)
\(y= sqrt((-2x+c_1)/(x^2+2x+1))\)
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