Differential equation with substitution

**Glitch** · January 4th 2011, 07:38 PM

The question:

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

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

given that y = 1 at x = 1.

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

Substituting initial value gives C = 3/2

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

**dwsmith** · January 4th 2011, 07:41 PM

$\displaystyle f(1,1)=\frac{1}{2}*1^2-1+c\Rightarrow c=\frac{1}{2}$

**Jhevon** · January 4th 2011, 07:44 PM

Originally Posted by Glitch

The question:

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

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

given that y = 1 at x = 1.

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

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

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

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

You must substitute that for dy/dx into your differential equation then proceed via separation of variables.

**Prove It** · January 4th 2011, 07:45 PM

No, it's not. You need to get a DE that does not have any reference to the variable $\displaystyle y$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

**Random Variable** · January 4th 2011, 07:46 PM

$\displaystyle y = vx$ which implies $\displaystyle \frac{dy}{dx} = x\frac{dv}{dx} + v$ by the chain rule (and product rule).

**Chris L T521** · January 4th 2011, 07:46 PM

Originally Posted by Glitch

The question:

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

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

given that y = 1 at x = 1.

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

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. $v=\frac{y}{x}$ . So now we need to rewrite $\dfrac{\,dy}{\,dx}$ :

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

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

Can you proceed?

EDIT: Man, am I slow!!

**nehme007** · January 4th 2011, 07:49 PM

You are integrating with respect to v when you should be integrating with respect to x. You need to make the substitution $\frac{dy}{dx} = v + x \frac{dv}{dx}$ and solve the resulting equation for v using separation of variables. Once you have v you can get y since y = vx.

**Glitch** · January 4th 2011, 11:00 PM

Thanks guys!

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

**Prove It** · January 5th 2011, 03:42 AM

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 $\displaystyle 3C \neq C$ , but yes, it's still just an arbitrary constant.

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