Integrating factor

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In the case of a linear first order homogeneous ODE such as au'(x)+bu(x)+c=0,\,\,u(0)=n\, the equation should always be divided by a so the coefficient of the u' term is unity.

u'+{\frac  {b}{a}}u+{\frac  {c}{a}}=0\,

The integrating factor (\rho ) is found by exponentiating the integral of the coefficient of the u term with respect to x. The integral is indefinate and you don't have to worry about constants of integration.

\rho =e^{{\int {\frac  {b}{a}}dx}}=e^{{{\frac  {b}{a}}x}}\,

Now, multiply the equation by the integrating factor.

e^{{{\frac  {b}{a}}x}}u'+{\frac  {b}{a}}e^{{{\frac  {b}{a}}x}}u+{\frac  {c}{a}}e^{{{\frac  {b}{a}}x}}=0\,

Notice that the first two terms of the left hand side (LHS) of the equation are the chain rule expansion of

{\frac  {d}{dx}}\left[e^{{{\frac  {b}{a}}x}}u\right]=-{\frac  {c}{a}}e^{{{\frac  {b}{a}}x}}\,

Integrate both sides with respect to x, and here constants of integration are required again.

e^{{{\frac  {b}{a}}x}}u=-{\frac  {c}{b}}e^{{{\frac  {b}{a}}x}}+c_{1}\,

u(x)=-{\frac  {c}{b}}+c_{1}e^{{-{\frac  {b}{a}}x}}\,

Plugging in 0 for the IC,

u(0)=-{\frac  {c}{b}}+c_{1}=n\,

c_{1}={\frac  {c}{b}}+n\,

So the unique solution for this IC is

u(x)=-{\frac  {c}{b}}+\left({\frac  {c}{b}}+n\right)e^{{-{\frac  {b}{a}}x}}\,

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