1st order ODE, now seperable

[willmoore21]

Ok so I have transformed a
1st order homogenous ODE into a seperable ODE. However I am having trouble seperating to get y on it's own.

Here's the problem:

\(\displaystyle du/dx=(2u^2)/x where u = y/x\)

du/(2u^2)=dx/x <<can't get tex to work

-1/(4u^2)=ln(x)+C=ln(Ax) <<can't get tex to work

1=-4u^2ln(Ax)

1=-4(y^2/x^2)ln(Ax)

y^2=-4x^2ln(Ax)

y=i2xsqrt(lnAx)

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Is this algebra correct? Is this the right solution? I'm not sure about bringing the y^2 over to the left is ok.






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[soroban]

Hello, willmoore21!

Your integration is off . . .

I have transformed a 1st order homogenous ODE into a seperable ODE.
However I am having trouble seperating to get y on it's own.

Here's the problem: .\(\displaystyle \dfrac{du}{dx}\:=\:\dfrac{2u^2}{x}\;\text{ where }u = \dfrac{y}{x}\)

We have: .\(\displaystyle u^{-2}\,du \:=\:\dfrac{2\,dx}{x}\)

Integrate: .\(\displaystyle -u^{-1} \;=\;2\ln(x) + K \;=\;\ln(x^2) + \ln(C) \;=\; \ln(Cx^2) \)

. . . . . . . . .\(\displaystyle -\dfrac{1}{u} \:=\:\ln(Cx^2) \quad\Rightarrow\quad u \:=\:\dfrac{-1}{\ln(Cx^2)} \)

. . . . . . . . .\(\displaystyle \dfrac{y}{x} \;=\;\dfrac{-1}{\ln(Cx^2)} \quad\Rightarrow\quad y \;=\;\dfrac{-x}{\ln(Cx^2)} \)​

 

[willmoore21]

Hello, willmoore21!

Your integration is off . . .

We have: .\(\displaystyle u^{-2}\,du \:=\:\dfrac{2\,dx}{x}\)

Integrate: .\(\displaystyle -u^{-1} \;=\;2\ln(x) + K \;=\;\ln(x^2) + \ln(C) \;=\; \ln(Cx^2) \)

. . . . . . . . .\(\displaystyle -\dfrac{1}{u} \:=\:\ln(Cx^2) \quad\Rightarrow\quad u \:=\:\dfrac{-1}{\ln(Cx^2)} \)

. . . . . . . . .\(\displaystyle \dfrac{y}{x} \;=\;\dfrac{-1}{\ln(Cx^2)} \quad\Rightarrow\quad y \;=\;\dfrac{-x}{\ln(Cx^2)} \)​

This is pretty funny error, silly mistakes. Thanks Soroban.