Respuesta :

syed514
(1−2x^2−2y)y′=4x^3+4xy

(1−2x^2−2y)dy=(4x^3+4xy)dx

(4x^3+4xy)dx+(−1+2x^2+2y)dy=0

f(x,y)=C;df(x,y)=∂f/∂x(dx)+∂f/∂y(dy)=0

∂f /∂x=4x^3+4xy⇒f(x,y)=x^4+2x^2y+g(y)

∂f/∂y=−1+2x^2+2y=2x^2+∂g/∂y⇒∂g/∂y=2y−1

g(y)=y^2−y⇒f(x,y)=x^4+2x^2y+y^2−y=(x^2+y)2−y=C
lublana

Answer:

[tex]u=-x^4-2x^2y+y-y^2[/tex]

Step-by-step explanation:

We are given that

[tex](1-2x^2-2y)\frac{dy}{dx}=4x^3+4xy[/tex]

We have to solve the given differential equation

[tex](1-2x^2-2y)dy=(4x^3+4xy)dx[/tex]

[tex](1-2x^2-2y)dy-(4x^3+4xy)dx=0[/tex]

Compare with [tex]Mdx+ndy=0[/tex]

Then, we get [tex]M=-(4x^3+4xy),N=(1-2x^2-2y)[/tex]

Exact differential equation

[tex]M_y=N_x[/tex]

[tex]M_y=-4x[/tex]

[tex]N_x=-4x[/tex]

[tex]M_y=N_x[/tex]

Hence, the differential equation is an exact differential equation.

Solution of exact differential is given by

[tex]u=\int M(x,y)dx+K(y)[/tex] where K(y) is a function of y.

[tex]u=\int -(4x^3+4xy) dx+k(y)[/tex] y treated as constant

[tex]u=-x^4-2x^2y+k(y)[/tex]

[tex]u_y=N[/tex]

[tex]-2x^2+k'(y)=1-2x^2-2y[/tex]

[tex]K'(y)=1-2y[/tex]

[tex]k(y)=y-y^2[/tex]

Substitute the value then we get

Then, [tex]u=-x^4-2x^2y+y-y^2[/tex]

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