Differential Equations

[jsbeckton]

I'm studying my old tests for finals and have come across on that I got wrong and can't recall how to do it.

Show that the differential equation:

$\displaystyle 2x{\rm }\ln {\rm y dx + }\left( {\frac{{x^2 }}{y}{\rm - }\sqrt[{}]{{\rm y}}} \right){\rm dy = 0}$

I've shown that $\displaystyle {\rm M}_{\rm y} {\rm = N}_{\rm x} {\rm } \Rightarrow {\rm }\frac{{{\rm 2x}}}{{\rm y}}{\rm = }\frac{{{\rm 2x}}}{{\rm y}}{\rm } \Rightarrow {\rm equality} \Rightarrow {\rm exact}$

I'm having trouble remembering how to get the general solution:

$\displaystyle x^2 {\rm ln y - }\frac{{\rm 2}}{{\rm 3}}{\rm y}^{\frac{{\rm 3}}{{\rm 2}}} {\rm = c}$

I've started with this:

\(\displaystyle \begin{array}{l}
g(x,y) = \int {2x{\rm }\ln y{\rm }dx} {\rm + c(y)} \\
{\rm = lny x}^{\rm 2} {\rm + c(y)} \\
\end{array}\)


Anyone know how to get this? Thanks.

 

[soroban]

Hello, jsbeckton!

I'm studying my old tests for finals and have come across on that I got wrong and can't recall how to do it.

Show that the differential equation:

\(\displaystyle 2x\cdot\ln y\cdot dx\,+\,\left( {\frac{x^2 }{y}\,-\,\sqrt{y}\right)dy\;=\;0\)

I've shown that: $\displaystyle M_y\,=\,N_x\;\;\Rightarrow \;\;\frac{2x}{y}\,=\;\frac{2x}{y}\;\;\Rightarrow\;\;\text{equality }\;\Rightarrow\;\text{ exact}$

I'm having trouble remembering how to get the general solution: $\displaystyle \;x^2\cdot\ln y\,-\,\frac{2}{3}y^{\frac{3}{2}}\:=\;C$

I've started with this:

$\displaystyle g(x,y)\;=\;\int M\,dx\;=\;\int 2x\cdot \ln y\,dx\;=\;x^2\cdot\ln y\,+\,c_1(y)$ . . . . You're doing fine!

Now integrate $\displaystyle N$ with respect to $\displaystyle y$.

\(\displaystyle \L g(x,y)\;=\;\int\left(\frac{x^2}{y}\,-\,y^{\frac{1}{2}}\right)\,dy\;=\;x^2\cdot\ln y\,-\,\frac{2}{3}y^{\frac{3}{2}}\,+\,c_2(x)\)


Now "combine" them: \(\displaystyle \L\;x^2\cdot\ln y\,-\,\frac{2}{3}y^{\frac{3}{2}} \;= \;C\)

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By the way, click on "Quote" and see the way I use LaTex.
. . It should have you hundreds of key-strokes.

 

[galactus]

See if this rings a bell.

$\displaystyle M(x,y)=2xlnydx$

$\displaystyle N(x,y)=(\frac{x^{2}}{y}-sqrt{y})dy$

$\displaystyle \frac{\partial{M}}{\partial{y}}=\frac{2x}{y}=\frac{\partial{N}}{\partial{x}}$

Exact, as you said.

$\displaystyle \frac{\partial{f}}{\partial{x}}=2xlny$ and $\displaystyle \frac{\partial{f}}{\partial{y}}=\frac{x^{2}}{y}-sqrt{y}$

$\displaystyle \int_(2xlny)dx=x^{2}lny+g(y)$

$\displaystyle \frac{\partial{f}}{\partial{y}}=\frac{d(x^{2}lny+g(y))}{dy}=\frac{x^{2}}{y}+g'(y)$

$\displaystyle g'{y}=-sqrt{y}$ and $\displaystyle g(y)=\frac{-2y^{\frac{3}{2}}}{3}$

$\displaystyle x^{2}lny+g(y)=x^{2}lny-\frac{2y^{\frac{3}{2}}}{3}=C$

 

[jsbeckton]

Thank you both, and thankfully bells are ringing

 

[jsbeckton]

do you know a better way to use LaTex or is that just your signature?