Initial Value Problem General Question

• Jan 14th 2009, 07:14 AM
rualin
Initial Value Problem General Question
Given an initial value problem with separable differential equation $\displaystyle \frac{dy}{dx}=\frac{g(x)}{h(y)}$ and initial condition $\displaystyle (x_0,y_0)$ is it true that $\displaystyle \int_{y_0}^y h(t)dt=\int_{x_0}^x g(t)dt$ and is that the same as solving by using indefinite integrals and then using the initial condition to solve for the constant? If yes, how? I'm having trouble understanding. Thank you!
• Jan 14th 2009, 07:47 AM
HallsofIvy
Quote:

Originally Posted by rualin
Given an initial value problem with separable differential equation $\displaystyle \frac{dy}{dx}=\frac{g(x)}{h(y)}$ and initial condition $\displaystyle (x_0,y_0)$ is it true that $\displaystyle \int_{y_0}^y h(t)dt=\int_{x_0}^x g(t)dt$ and is that the same as solving by using indefinite integrals and then using the initial condition to solve for the constant? If yes, how? I'm having trouble understanding. Thank you!

Yes, they are the same.

Let F(x) be an anti-derivative of f(x) and G(t) be an anti-derivative of g(t).
Then $\displaystyle \int_{y_0}^y f(t)dt= F(y)- F(y_0)$ and $\displaystyle \int_{x_0}^x g(t)dt= G(x)- G(x_0)$. Your solution becomes $\displaystyle F(y)- F(y_0)= G(x)- G(x_0)$ which can be written $\displaystyle F(y)= G(x)+ (F(y_0)- G(x_0))$.

Now, suppose you used indefinite integrals: $\displaystyle \int f(y)dy= F(y)+ C_1$ and $\displaystyle \int g(x)dx= G(x)+ C_2$ so you would have $\displaystyle F(y)+ C_1= G(x)+ C_2$ or just $\displaystyle F(y)= G(x)+ C$ where $\displaystyle C= C_2-C_1$. Now put $\displaystyle y= y_0$, $\displaystyle x= x_0$ into that: $\displaystyle F(y_0)= G(x_0)+ C$ so $\displaystyle C= F(y_0)- G(x_0)$. That gives $\displaystyle F(y)= G(x)+ (F(y_0)- G(x_0))$, exactly the same answer as above.
• Jan 14th 2009, 08:05 AM
rualin
Nice explanation! Thanks!