Solve the differential equation dy/dx + 2y = e^(3x).

Step-by-step solution

1. 1

   Confirm the standard first-order linear form. With $y'+P(x)y=Q(x)$ we read

   $P(x)=2,\qquad Q(x)=e^{3x}$

   Only $Q$ differs from the $e^{-x}$ version of this problem, so the machinery is identical and only the final integral changes.

2. 2

   Compute the integrating factor from $P$ alone. Because $\mu$ depends only on $P$, the forcing term $e^{3x}$ has no say here:

   $\mu(x)=e^{\int 2\,dx}=e^{2x}$

3. 3

   Multiply through so the left side becomes one derivative.

   $e^{2x}\frac{dy}{dx}+2e^{2x}y=e^{2x}e^{3x}=e^{5x}$

   $\frac{d}{dx}\left(e^{2x}y\right)=e^{5x}$

   Note the exponents add: this is why the right side is $e^{5x}$ and not $e^{3x}$.

4. 4

   Integrate, using $\int e^{kx}dx=e^{kx}/k$. Here $k=5$, so the $1/5$ that ends up in the final answer is born at this step:

   $e^{2x}y=\frac{1}{5}e^{5x}+C$

5. 5

   Divide by $e^{2x}$ and simplify the exponents.

   $y=\frac{1}{5}e^{5x-2x}+Ce^{-2x}=\frac{1}{5}e^{3x}+Ce^{-2x}$

6. 6

   Check by direct substitution. From $y=\tfrac15 e^{3x}+Ce^{-2x}$, $y'=\tfrac35 e^{3x}-2Ce^{-2x}$, so

   $y'+2y=\left(\tfrac35+\tfrac25\right)e^{3x}+(-2C+2C)e^{-2x}=e^{3x}$

   The coefficient check $\tfrac35+\tfrac25=1$ is the part worth doing by hand — a wrong $1/5$ would show up immediately.