Integrating Factor

When solving first order linear differential equations of the form $y'+p(x)y=g(x)$, we can utilize the "integrating factor" $\mu (x)=e^{\int p(x)dx}$.

Steps to solving an equation of the form ${\frac {dy}{dx}}+p(x)y=g(x)$:

1. Find the integrating factor $\mu (x)=e^{\int p(x)dx}$, and note that $\mu '(x)=p(x)e^{\int p(x)dx}=p(x)\mu (x)$,
2. Multiply both sides of the equation by the integrating factor.
3. The left side of the equation, $y'\mu (x)+p(x)\mu (x)y$, can now be rewritten as $(\mu (x)y)'$ since $y'\mu (x)+p(x)\mu (x)y=y'\mu (x)+\mu '(x)y$. Verify by taking the derivative of $\mu (x)y$ with respect to x with the product rule.
4. Now, integrate $(\mu (x)y)'=g(x)\mu (x)$ to get $\mu (x)y=\int g(x)\mu (x)dx$.
5. Solve for y.

Example problem: $y'+{\frac {2}{t}}y=t-1+{\frac {1}{t}}$

1. $\mu (x)=e^{\int {\frac {2}{t}}dt}=e^{2\ln {|t|}}=e^{\ln {|t|^{2}}}=t^{2}$
2. $t^{2}y'+2ty=t^{3}-t^{2}+t$
3. $(t^{2}y)'=t^{3}-t^{2}+t$
4. $t^{2}y=\int (t^{3}-t^{2}+t)dt={\frac {1}{4}}t^{4}-{\frac {1}{3}}t^{3}+{\frac {1}{2}}t^{2}+C$
5. $y={\frac {1}{4}}t^{2}-{\frac {1}{3}}t+{\frac {1}{2}}+{\frac {C}{t^{2}}}$

Resources

• Solving Linear Equations, Paul's Online Notes