# Differential equation with substitution

• Jan 4th 2011, 07:38 PM
Glitch
Differential equation with substitution
The question:

Use the substitution $\displaystyle v = \frac{y}{x}$ to solve

$\displaystyle \frac{dy}{dx} = \frac{(\frac{y}{x})^3 + 1}{(\frac{y}{x})^2}$

given that y = 1 at x = 1.

My attempt:
$\displaystyle \frac{dy}{dx} = \frac{v^3 + 1}{v^2}$
$\displaystyle y = \int{\frac{v^3}{v^2}} + \int{\frac{1}{v^2}}$
= $\displaystyle \int{v} + \int{\frac{1}{v^2}}$
= $\displaystyle \frac{v^2}{2} - \frac{1}{v} + C$
= $\displaystyle \frac{1}{2}(\frac{y}{x})^2} - \frac{1}{\frac{y}{x}} + C$
= $\displaystyle \frac{1}{2}(\frac{y}{x})^2} - \frac{x}{y} + C$

Substituting initial value gives C = 3/2

Is this correct? I don't have answers for practice exams. :(
• Jan 4th 2011, 07:41 PM
dwsmith
$\displaystyle \displaystyle f(1,1)=\frac{1}{2}*1^2-1+c\Rightarrow c=\frac{1}{2}$
• Jan 4th 2011, 07:44 PM
Jhevon
Quote:

Originally Posted by Glitch
The question:

Use the substitution $\displaystyle v = \frac{y}{x}$ to solve

$\displaystyle \frac{dy}{dx} = \frac{(\frac{y}{x})^3 + 1}{(\frac{y}{x})^2}$

given that y = 1 at x = 1.

My attempt:
$\displaystyle \frac{dy}{dx} = \frac{v^3 + 1}{v^2}$
$\displaystyle y = \int{\frac{v^3}{v^2}} + \int{\frac{1}{v^2}}$
= $\displaystyle \int{v} + \int{\frac{1}{v^2}}$
= $\displaystyle \frac{v^2}{2} - \frac{1}{v} + C$
= $\displaystyle \frac{1}{2}(\frac{y}{x})^2} - \frac{1}{\frac{y}{x}} + C$
= $\displaystyle \frac{1}{2}(\frac{y}{x})^2} - \frac{x}{y} + C$

Substituting initial value gives C = 3/2

Is this correct? I don't have answers for practice exams. :(

No, that's not correct. You tried to solve the problem with 3 variables. You integrated both sides with respect to x while treating v as the variable too. absurdity :p

$\displaystyle \displaystyle v = \frac yx \implies y = vx$

$\displaystyle \displaystyle \Rightarrow \frac {dy}{dx} = \frac {dv}{dx} \cdot x + v$

You must substitute that for dy/dx into your differential equation then proceed via separation of variables.
• Jan 4th 2011, 07:45 PM
Prove It
No, it's not. You need to get a DE that does not have any reference to the variable $\displaystyle \displaystyle y$.

$\displaystyle \displaystyle \frac{dy}{dx} = \frac{\left(\frac{y}{x}\right)^3 + 1}{\left(\frac{y}{x}\right)^2}$.

Make the substitution $\displaystyle \displaystyle v = \frac{y}{x}$. Then $\displaystyle \displaystyle y = v\,x$ and $\displaystyle \displaystyle \frac{dy}{dx} = x\,\frac{dv}{dx} + v$ and the DE becomes

$\displaystyle \displaystyle x\,\frac{dv}{dx} + v = \frac{v^3 + 1}{v^2}$

$\displaystyle \displaystyle x\,\frac{dv}{dx} + v = v + \frac{1}{v^2}$

$\displaystyle \displaystyle x\,\frac{dv}{dx} = \frac{1}{v^2}$

$\displaystyle \displaystyle v^2\,\frac{dv}{dx} = \frac{1}{x}$

$\displaystyle \displaystyle \int{v^2\,\frac{dv}{dx}\,dx} = \int{\frac{1}{x}\,dx}$

$\displaystyle \displaystyle \int{v^2\,dv} = \ln{|x|} + C_1$

$\displaystyle \displaystyle \frac{v^3}{3} + C_2 = \ln{|x|} + C_1$

$\displaystyle \displaystyle \frac{v^3}{3} = \ln{|x|} + C$ where $\displaystyle \displaystyle C = C_1-C_2$

$\displaystyle \displaystyle v^3 = 3\ln{|x|} + 3C$

$\displaystyle \displaystyle v = \sqrt[3]{3\ln{|x|} + 3C}$

$\displaystyle \displaystyle \frac{y}{x} = \sqrt[3]{3\ln{|x|} + 3C}$

$\displaystyle \displaystyle y = x\sqrt[3]{3\ln{|x|} + 3C}$.
• Jan 4th 2011, 07:46 PM
Random Variable
$\displaystyle \displaystyle y = vx$ which implies $\displaystyle \displaystyle \frac{dy}{dx} = x\frac{dv}{dx} + v$ by the chain rule (and product rule).
• Jan 4th 2011, 07:46 PM
Chris L T521
Quote:

Originally Posted by Glitch
The question:

Use the substitution $\displaystyle v = \frac{y}{x}$ to solve

$\displaystyle \frac{dy}{dx} = \frac{(\frac{y}{x})^3 + 1}{(\frac{y}{x})^2}$

given that y = 1 at x = 1.

My attempt:
$\displaystyle \frac{dy}{dx} = \frac{v^3 + 1}{v^2}$
$\displaystyle y = \int{\frac{v^3}{v^2}} + \int{\frac{1}{v^2}}$
= $\displaystyle \int{v} + \int{\frac{1}{v^2}}$
= $\displaystyle \frac{v^2}{2} - \frac{1}{v} + C$
= $\displaystyle \frac{1}{2}(\frac{y}{x})^2} - \frac{1}{\frac{y}{x}} + C$
= $\displaystyle \frac{1}{2}(\frac{y}{x})^2} - \frac{x}{y} + C$

Substituting initial value gives C = 3/2

Is this correct? I don't have answers for practice exams. :(

I'm sorry to say that this is not correct, because you're integrating with respect to the wrong variable (v instead of x).

First apply the substitution that's given to you. $\displaystyle v=\frac{y}{x}$. So now we need to rewrite $\displaystyle \dfrac{\,dy}{\,dx}$:

$\displaystyle \dfrac{\,dv}{\,dx}=-\dfrac{y}{x^2}+\dfrac{1}{x}\dfrac{\,dy}{\,dx}$. Since $\displaystyle y=vx$, we end up with $\displaystyle \dfrac{\,dy}{\,dx}=v+x\dfrac{\,dv}{\,dx}$

Thus, the ODE becomes $\displaystyle v+x\dfrac{\,dv}{\,dx}=\dfrac{v^3+1}{v^2}$. This is now a separable equation.

Can you proceed?

EDIT: Man, am I slow!!
• Jan 4th 2011, 07:49 PM
nehme007
You are integrating with respect to v when you should be integrating with respect to x. You need to make the substitution $\displaystyle \frac{dy}{dx} = v + x \frac{dv}{dx}$ and solve the resulting equation for v using separation of variables. Once you have v you can get y since y = vx.
• Jan 4th 2011, 11:00 PM
Glitch
Thanks guys!

With the 3C in the solution, is it fair to just write 'C' since it's still an unknown constant?
• Jan 5th 2011, 03:42 AM
Prove It
Quote:

Originally Posted by Glitch
Thanks guys!

With the 3C in the solution, is it fair to just write 'C' since it's still an unknown constant?

I'd use a different letter just because $\displaystyle \displaystyle 3C \neq C$, but yes, it's still just an arbitrary constant.