# variables separable ODE

• Apr 7th 2008, 04:11 AM
johnbarkwith
variables separable ODE
When solving a variables separable ODE such as:

dy/dx=f(x) => intergral dy = integral f(x) dx

what limits does one integrate between??????
• Apr 7th 2008, 05:13 AM
mr fantastic
Quote:

Originally Posted by johnbarkwith
When solving a variables separable ODE such as:

dy/dx=f(x) => intergral dy = integral f(x) dx

what limits does one integrate between??????

If the boundary condition $\displaystyle f(x_1) = y_1$ is given, then

$\displaystyle y = \int_{x_1}^{x} f(w) \, dw + y_1$.

If the boundary condition $\displaystyle f(x_1) = y_1$ is given and the the value of y when $\displaystyle x = x_2$ is required, then

$\displaystyle y = \int_{x_1}^{x_2} f(w) \, dw + y_1$.

Otherwise there are no limits:

$\displaystyle y = \int f(x) \, dx$

and the right hand side is an indefinite integral. Of course, this approach can also be taken when boundary conditions are given - see the next reply.
• Apr 7th 2008, 05:13 AM
awkward
Usually you don't use limits on the integral. You evaluate the indefinite integral and add in a constant of integration. Then if you have an initial value for your differential equation you use it to find the value of the constant.
• Apr 7th 2008, 09:41 AM
Mathstud28
Maybe
this sample example will help $\displaystyle \frac{dy}{dx}=9y^4$...seperating we get $\displaystyle \frac{dy}{y^4}=9dx$ then by integrating we get $\displaystyle \frac{-1}{3y^3}=9x+C$ multiplying both sides by $\displaystyle -3$ we get $\displaystyle \frac{1}{y^3}=-27x+C_1$ and using the rules of ratios we can switch the means to give us $\displaystyle y^3=\frac{1}{-27x+c_1}$ finally by taking the cuberoot of each side we arrive at our answer $\displaystyle y=\frac{1}{[-27x+C_1]^{ \frac{1}{3}}}$