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Solve u'+3u=0,\,\,u(0)=2\,

Find the integrating factor \rho =e^{{\int 3dx}}=e^{{3x}}\,

Multiply it through the original equation:


Notice the left hand side (LHS) of the last equation implies that

{\frac  {d}{dx}}\left[e^{{3x}}u\right]=0\,

Integrate wrt x:

e^{{3x}}u=c\, where c is a constant.

u(x)=ce^{{-3x}}\, is the homogeneous solution. Plug in the IC to get:

u(0)=c=2\,. So the unique solution that satisfies this IC is:


Ordinary Differential Equations

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