Answer:
\(u=-x^4-2x^2y+y-y^2\)
Step-by-step explanation:
We are given that
\((1-2x^2-2y)\frac{dy}{dx}=4x^3+4xy\)
We have to solve the given differential equation
\((1-2x^2-2y)dy=(4x^3+4xy)dx\)
\((1-2x^2-2y)dy-(4x^3+4xy)dx=0\)
Compare with \(Mdx+ndy=0\)
Then, we get \(M=-(4x^3+4xy),N=(1-2x^2-2y)\)
Exact differential equation
\(M_y=N_x\)
\(M_y=-4x\)
\(N_x=-4x\)
\(M_y=N_x\)
Hence, the differential equation is an exact differential equation.
Solution of exact differential is given by
\(u=\int M(x,y)dx+K(y)\) where K(y) is a function of y.
\(u=\int -(4x^3+4xy) dx+k(y)\) y treated as constant
\(u=-x^4-2x^2y+k(y)\)
\(u_y=N\)
\(-2x^2+k'(y)=1-2x^2-2y\)
\(K'(y)=1-2y\)
\(k(y)=y-y^2\)
Substitute the value then we get
Then, \(u=-x^4-2x^2y+y-y^2\)
