# Math Help - First-order Separable ODE

1. ## First-order Separable ODE

Hi
Can you tell me if i am correct in my method?
thanks
John

Separate the variables and find the indefinite integrals.

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

re-arrange to get y on left hand side and x on right hand side

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

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

Now integrate both sides:

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

$lny=\frac{x^4}{4}-\frac{2}{3}x^\frac{3}{2} +C$

$y=e^{\frac{x^4}{4}-\frac{2}{3}x^\frac{3}{2} +C }$

$y= e^{\frac{x^4}{4}}-e^{\frac{2}{3}x^\frac{3}{2}}+e^{c}$

As $e^C=C$

$y= e^{\frac{x^4}{4}}-e^{\frac{2}{3}x^\frac{3}{2}}+C$

$y=e^{\frac{x^4}{4}-\frac{2}{3}x^\frac{3}{2}} +C$

Can you tell me if i am on track?

2. ## Re: Differential equation

If you want to check whether your answer is correct, you can always substitute your solution in the original differential equation.

I can save you some time in this instance though. Your work is correct up until the following:

$y = e^{\frac{x^4}{4}}-e^{\frac{2}{3}x^\frac{3}{2}}+e^C$

This is incorrect. Do you see why?

3. ## Re: Differential equation

Originally Posted by harbottle
If you want to check whether your answer is correct, you can always substitute your solution in the original differential equation.

I can save you some time in this instance though. Your work is correct up until the following:

$y = e^{\frac{x^4}{4}}-e^{\frac{2}{3}x^\frac{3}{2}}+e^C$

This is incorrect. Do you see why?
No i don't see it?

I have the $ln$ function on the left and to bring it across i get the inverse $Ln$ or $e$ function of the right hand side?

Can you tell me what's incorrect?

4. ## Re: Differential equation

You have $y = e^{\frac{x^4}{4} - \frac{2}{3}x^{\frac{3}{2}}+C}$, which is correct.

It does not follow that $y = e^{\frac{x^4}{4}} - e^{ \frac{2}{3}x^{\frac{3}{2}}} + e^C$, because that's not how exponents work. Instead the next line is:

$y = e^C\cdot e^{\frac{x^4}{4} - \frac{2}{3}x^{\frac{3}{2}}}$

5. ## Re: Differential equation

I need to work on my simplifying of exponents,

An example on another site used the same steps as i used above(which is wrong in this case). In that case the expression was a polynomial-maybe thats the difference?

thanks
John

6. ## Re: First-order Separable ODE

Originally Posted by celtic1234
Hi
Can you tell me if i am correct in my method?
thanks
John

Separate the variables and find the indefinite integrals.

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

re-arrange to get y on left hand side and x on right hand side

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

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

Now integrate both sides:

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

$lny=\frac{x^4}{4}-\frac{2}{3}x^\frac{3}{2} +C$

$y=e^{\frac{x^4}{4}-\frac{2}{3}x^\frac{3}{2} +C }$

$y= e^{\frac{x^4}{4}}-e^{\frac{2}{3}x^\frac{3}{2}}+e^{c}$

As $e^C=C$

$y= e^{\frac{x^4}{4}}-e^{\frac{2}{3}x^\frac{3}{2}}+C$

$y=e^{\frac{x^4}{4}-\frac{2}{3}x^\frac{3}{2}} +C$

Can you tell me if i am on track?
First of all, $\displaystyle \int{\frac{1}{y}\,dy} = \ln{|y|} + C_1$, not $\displaystyle \ln{y}$...