Question

need help with 2 questions differential equations:
1) y' = (2x-y^2) / (y-2xy)

2) (x^3)y' = 2(y^2 + (x^2)y - x^4)

Answer 1

Try using the app mathway, it's a good calculator for problems like this and is super easy to use :)

Answer 2

i checked it out, but that doesn't even do differential equations....

Answer 3

Have you solved exact equations before?

Answer 4

yes, i tried the method but My and Nx are not equal to each other or am i wrong?

Answer 5

Have you tried finding an integrating factor to make them exact?

Answer 6

i tried, but i became too complex

Answer 7

hello, i believe i got number 1, any help in number two would be appreciated.

Answer 8

\[\begin{align*}x^3y'&=2(y^2+x^2y-x^4)\\\\
\frac{1}{y^2}y'&=\frac{2}{x^3}+\frac{2}{xy}-2x
\end{align*}\]
Let \(t=\dfrac{1}{y}\), so that \(t'=-\dfrac{1}{y^2}y'\).
\[\begin{align*}-t'&=\frac{2}{x^3}+\frac{2}{x}t-2x\\\\
t'-\frac{2}{x}t&=2x-\frac{2}{x^3}
\end{align*}\]
which is linear in \(t\).

Answer 9

As for the first problem, using the exact equation method,
\[\begin{align*}
y'&=\frac{2x-y^2}{y-2xy}\\\\
y^2-2x+(y-2xy)y'&=0
\end{align*}\]
Partial derivatives:
\[M_y=2y\\
N_x=-2y\]
Integrating factor:
\[\begin{align*}\ln\mu(x)&=\int\frac{M_y-N_x}{N}\,dx\\\\&=\int\frac{2y-(-2y)}{y-2xy}\,dx\\\\&=\int\frac{4}{1-2x}\,dx\\\\
&=-2\ln(1-2x)\\\\
\mu(x)&=\frac{1}{(1-2x)^2}
\end{align*}\]
So the ODE is now
\[\frac{1}{(1-2x)^2}(y^2-2x)+\frac{1}{(1-2x)^2}(y-2xy)y'=0\]
with partial derivatives,
\[{M^*}_y=\frac{2y}{(1-2x)^2}\\
{N^*}_x=\frac{2y}{(1-2x)^2}\]

Answer 10

thanks a lot!

Answer 11

yw

Answer 12

A simpler way for solving the first equation involves a substitution, \(y=\sqrt z\), which gives \(y^2=z\) and \(y'=\dfrac{z'}{2\sqrt z}\).
\[\begin{align*}
y'&=\frac{2x-y^2}{y-2xy}\\\\
\frac{1}{2\sqrt z}z'&=\frac{2x-z}{\sqrt z(1-2x)}\\\\
\frac{1}{2}z'&=\frac{2x-z}{1-2x}\\\\
z'+\frac{2}{1-2x}z&=\frac{4x}{1-2x}
\end{align*}\]
which is linear in \(z\).

Answer 13

do you mind looking at this, i used the exact method, i also used symbolab calculator for partial derivatives and integrals, Im not sure if its correct.

Answer 14

first picture might be small but you can ctrl + +, to zoom in

Answer 15

Also your second method is not correct, you second to last step to last step is not possible.

Answer 16

nvm its correct lol

Answer 17

In your work, your partial derivatives should match up. As above, you have
\[{M^*}=\frac{y^2-2x}{(1-2x)^2}~~\implies~~{M^*}_y=\frac{2y}{(1-2x)^2}\\
N^*=\frac{y-2xy}{(1-2x)^2}=\frac{y}{1-2x}~~\implies~~{N^*}_x=\frac{2y}{(1-2x)^2}\]
You're looking for a solution of the form \(\Psi(x,y)=C\), where \(\Psi_x=M^*\) and \(\Psi_y=N^*\dfrac{dy}{dx}\). Integrating, you have
\[\Psi=\int N^*\,dy=\int\frac{y}{1-2x}\,dy=\frac{y^2}{2(1-2x)}+f(x)\]
Differentiating with respect to \(x\) yields
\[\Psi_x=M^*=\frac{y^2-2x}{(1-2x)^2}=\frac{y^2}{(1-2x)^2}+f'(x)\]
which admits the ODE
\[f'(x)=-\frac{2x}{(1-2x)^2}\]
Integrating yields
\[f(x)=-2\int\frac{x}{(1-2x)^2}\,dx=-\frac{1}{2(1-2x)}-\frac{1}{2}\ln(1-2x)+C\]
and so the final solution should be
\[\Psi(x,y)=\frac{y^2}{2(1-2x)}-\frac{1}{2(1-2x)}-\frac{1}{2}\ln(1-2x)=C\]

Answer 18

This can of course be rewritten as needed.
\[\Psi(x,y)=\frac{y^2-1}{2(1-2x)}-\ln\sqrt{1-2x}=C\]
etc.

Answer 19

i already handed in the hw, but letting you know, the second to last step is incorrect, under integrating yields. integration and distribution is not correct.

Answer 20

anyways, ty very much, you are good with this.

Answer 21

Oh sorry, I must have missed that factor of \(x\) in the numerator...