# General Solution of a Differential Equation

• Mar 16th 2010, 08:00 PM
jaijay32
General Solution of a Differential Equation
Hi. The instructions say "Verify that y is a general solution of the differential equation. Then find the particular solution of the differential equation that satisfies the condition y(1) = -.5e".

Y= (Ce^x)/(x) + (.5x)(e^x).

Y' + [(1-X)/(X)]Y=e^x.

However, when I plug in the point, C always dissapears due to it being associated with a zero that is multiplied. The right answer is that
c= -1. How do I arrive at thiis answer?

Many thanks
• Mar 16th 2010, 08:22 PM
dedust
Quote:

Originally Posted by jaijay32
Hi. The instructions say "Verify that y is a general solution of the differential equation. Then find the particular solution of the differential equation that satisfies the condition y(1) = -.5e".

Y= (Ce^x)/(x) + (.5x)(e^x).

Y' + [(1-X)/(X)]Y=e^x.

However, when I plug in the point, C always dissapears due to it being associated with a zero that is multiplied. The right answer is that
c= -1. How do I arrive at thiis answer?

Many thanks

\$\displaystyle C=-1\$, and your particular solution is \$\displaystyle Y(x) = (-e^x)/(x) + (.5x)(e^x)\$
• Mar 17th 2010, 02:39 AM
HallsofIvy
Quote:

Originally Posted by jaijay32
Hi. The instructions say "Verify that y is a general solution of the differential equation. Then find the particular solution of the differential equation that satisfies the condition y(1) = -.5e".

Y= (Ce^x)/(x) + (.5x)(e^x).

Y' + [(1-X)/(X)]Y=e^x.

However, when I plug in the point, C always dissapears due to it being associated with a zero that is multiplied. The right answer is that
c= -1. How do I arrive at thiis answer?

Many thanks

If y= (Ce^x)/(x)+ (.5x)e^x then y(1)= (Ce^1)/(1)+ (.5(1))e^1= Ce+ .5e= -.5e. Subtracting .5e from both sides Ce= -e.