Question

Find the general solution of the equation
(y^2 cosx-3x^2y-2x)dx+(2ysinx-x^3+lny)dy=0

Answer 1

My 1st try is 2 show this diffenrential equation is an exact equation.
So,\[M =y ^{2}\cos x -3x ^{2}y -2x\]
\[\frac{ dM }{ dy }= 2ycox-3x ^{2}\]
\[N =2ysinx -x ^{3}+ \ln y\]
\[\frac{ dN}{ dx }=2y \cos x -3x ^{2}\]
Therefore the differential equation is an exact eqution.

Answer 2

Next,I integrate M and N respectively where
\[U{1}= y ^{2}\sin x -x ^{3}y -2xy +\phi(y)\]
\[U{2}=y ^{2}\sin x -x ^{3}y +\frac{ 1 }{ y}+\phi(x)\]
Then i compare u1 and u2 in order to get the \[\phi(y) and \phi(x)\]

Answer 3

So it become \[-2xy +\phi(y)=\frac{ 1 }{ y}+\phi(x)\]
Then, what is my x and y?

Answer 4

why dont you simply just, after your work,do this :
dM/dN * dx/dy = 1
or
dy/dx = dM/dN

your original eqn was
Ndy + Mdx =0
=> dy/dx = -M/N = dM/dN

simpler now ?

Answer 5

Yeah,its bcome simpler.Bt what ll b the ans if i wanna use the method i did above?

Answer 6

\[(y^2 cosx-3x^2y-2x)dx+(2ysinx-x^3+lny)dy=0\]
\[M=y^2 cosx-3x^2y-2x\]\[N=2ysinx-x^3+lny\]

\[\frac{\partial M}{\partial y}=2ycosx-3x^2\]\[\frac{\partial N}{\partial x}=2ycosx-3x^2=\frac{\partial M}{\partial y}\]So, the DE is exact.

f(x,y) = c is a solution
\[\frac{\partial f}{\partial x}=M=y^2 cosx-3x^2y-2x\]\[f(x,y) = \int (y^2 cosx-3x^2y-2x )dx=y^2sinx-x^3y-x^2+\phi (y)\]\[\frac{\partial f}{\partial y}=\frac{\partial }{\partial y}(y^2sinx-x^3y-x^2+\phi (y))=N\]\[2ysinx-x^3+\phi'(y)=2ysinx-x^3+lny\]\[\phi'(y)=lny\]So, you can get \(\phi '(y)\) from here.

Answer 7

Sorry! Last sentence: So, you can get ϕ(y) from here.

Answer 8

Then,
\[f(x,y) = y^2sinx-x^3y-x^2+\int lny dy=C\] (Do the integration..)

Answer 9

Hmm,i see my mistake.Thx ur guys' help :)