Advertisements

Why Muss with the “+C”?

Lin McMullin / June 21, 2013

Here is a way to find the particular solution of a separable differential equation without using the +C and it might even be a little faster. As an example consider the initial value problem

$\displaystyle \frac{dy}{dx}=\frac{y-1}{{{x}^{2}}}\text{ with }x\ne 0\text{ and }y\left( 2 \right)=0$

Separate the variables as usual and then write each side a definite integral with the lower limit of integration the number from the initial condition and the upper limit variable. I’ll replace the dummy variable in the integrand with an upper case X or Y to avoid having x and y in two places.

$\displaystyle \int_{0}^{y}{\frac{1}{Y-1}dY}=\int_{2}^{x}{{{X}^{-2}}dX}$

Integrate

$\displaystyle \left. \ln \left| Y-1 \right| \right|_{0}^{y}=\left. -{{X}^{-1}} \right|_{2}^{x}$

$\displaystyle \ln \left| y-1 \right|-\ln \left| 0-1 \right|=-\frac{1}{x}+\frac{1}{2}$

Note that near the initial condition where y = 0, $\left( y-1 \right)<0$ so $\left| y-1 \right|=-\left( y-1 \right)=1-y$ . Continue and solve for y.

$\displaystyle {{e}^{\ln \left( 1-y \right)}}={{e}^{-\frac{1}{x}+\frac{1}{2}}}$

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

$\displaystyle y=1-{{e}^{-\frac{1}{x}+\frac{1}{2}}},\quad x>0$

No muss; no fuss.

Advertisements

Leave a Reply Cancel reply

Enter your comment here...

Fill in your details below or click an icon to log in:

Email (required) (Address never made public)

Name (required)

Website

You are commenting using your WordPress.com account. ( Log Out / Change )

You are commenting using your Twitter account. ( Log Out / Change )

You are commenting using your Facebook account. ( Log Out / Change )

You are commenting using your Google+ account. ( Log Out / Change )

Connecting to %s

%d bloggers like this: