Question

solve the diffrential equation (x^4-2xy^2+y^4)dx-(2x^2y-4xy^3+ siny)dy =0​

Answer 1

Answer:

x^5/5 -x^2y^2+xy^4-cosy+C

Step-by-step explanation:

This is homogenous equation in order to began to solve it you must do the exactness differential equation test

which is found in this image

https://www.google.com/url?sa=i&source=images&cd=&ved=2ahUKEwiN35Ko28zlAhWldN8KHRlBAFoQjRx6BAgBEAQ&url=https%3A%2F%2Fslideplayer.com%2Fslide%2F8582157%2F&psig=AOvVaw1PWg_oFy3iPTAdZBsKusYi&ust=1572824851286584

as you can see that M(x,y) represents the side of equation attached to dx

N(x,y) represents the side of equation attached to dy

dx represents the differentiation respect to x

dy represents the differentiation respect to y

so the function was having both x and y as variables

F(x,y)

we call it F

when you differentiate it you began to

dF=dF/dx +dF/dy

so the problem is how to return it back

the test can solve this problem as it was proven by mathematician

This test you can understand its concept even more by watching this video

https://www.youtube.com/watch?v=WgHCy25hBJE&t=829s

now to do you the test you began to differentiate the side attached to dx with respect to y so you treat x as a constant

4xy-4y^3

you diiferentiate side attached to  dy with respect to x so you treat y as a  constant

4xy-4y^3

they are the same so you begin to solve the problem

by integrating the side attached to dx with respect to x

x^5/5 -x^2y^2+xy^4+k(y)

this k represent function of y as you differentiated x you removed y as it wads constant with respect to x

this is F(x,y) so you have to diferentiate it with respect to y in order to find k(y)

2xy^2+4xy^3+k'(y)

you equal it with the side attached to dy in the main equation

you will find that

k'(y)=siny

by integration

so k(y)=--cosy+C

you subsitute it you will get the answer

Answer 2

Answer:

4-2xy^2+y^4)dx-(2x^2y-4xy^3+ siny)dy =0