Seperable DE

**TrevorP** · April 13th 2008, 01:47 PM

Hey,

I've been having some trouble recreating the answer my teacher has for the following question:

Find $y(x)$ given $\frac{dy}{dx} = x^3y$ , where $y(1) = 2$ .

So I did the following:

$\int \frac{dy}{y} = \int x^3dx \longrightarrow \ln{y} = \frac{x^4}{4} + C$

$y = e^{\frac{x^4}{4} + C} \ \ or \ \ y = Ce^{\frac{x^4}{4}}$

Where C is some constant to be solved for.

Anyway I could solve for C. But what is getting me is that my prof's final answer is:

$y = 2e^{\frac{x^4}{4} - \frac{1}{4}}$

Which would mean there was two constants...but I would need more initial conditions for that.

**Mathstud28** · April 13th 2008, 01:55 PM

Originally Posted by TrevorP

Hey,

I've been having some trouble recreating the answer my teacher has for the following question:

Find $y(x)$ given $\frac{dy}{dx} = x^3y$ , where $y(1) = 2$ .

So I did the following:

$\int \frac{dy}{y} = \int x^3dx \longrightarrow \ln{y} = \frac{x^4}{4} + C$

$y = e^{\frac{x^4}{4} + C} \ \ or \ \ y = Ce^{\frac{x^4}{4}}$

Where C is some constant to be solved for.

Anyway I could solve for C. But what is getting me is that my prof's final answer is:

$y = 2e^{\frac{x^4}{4} - \frac{1}{4}}$

Which would mean there was two constants...but I would need more initial conditions for that.

You have that $2=Ce^{\frac{1^4}{4}}$ dividing each side we get $c=\frac{2}{e^{\frac{1}{4}}}\Rightarrow2e^{\frac{-1}{4}}$ see it now?

**Soroban** · April 13th 2008, 02:21 PM

Hello, Trevor!

You're both correct . . .

$y \:= \:e^{\frac{x^4}{4} + C}\:\text{ or } \:y \:= \:Ce^{\frac{x^4}{4}}$ .where $C$ is some constant to be solved for.

I could solve for C. . . . . Did you?

$\text{My prof's final answer is: }\;y \:= \:2\,e^{\frac{x^4}{4} - \frac{1}{4}}$

Since $y(1) = 2$ , we have: . $Ce^{\frac{1}{4}} \:=\:2\quad\Rightarrow\quad C \:=\:2e^{\text{-}\frac{1}{4}}$

Then the function becomes: . $y \;=\;2\underbrace{e^{\text{-}\frac{1}{4}}\cdot e^{\frac{x^2}{4}}}_{\text{combine}} \;=\;2\,e^{ \frac{x^2}{4}-\frac{1}{4}}$ . . . see?

**TrevorP** · April 13th 2008, 02:41 PM

Yeah I did..but something just wasn't clicking for some reason. I just thought that since two numbers were introduced there should be two variables.

But now it makes sense.

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