Solve a first-order linear equation with an integrating factor
Write the equation in the form y′ + P(x)y = Q(x) before choosing the integrating factor μ = e^(∫P dx). Multiplying every term by μ makes the left side the derivative of μy. Integrate that product equation, divide by μ and only then apply the initial condition. A separable equation and a linear equation are classifications that can overlap.
Worked practice example
Solve y′ + 2y = 6 with y(0) = 1.
| Step | Working |
|---|---|
| Integrating factor | P = 2, so μ = e^(2x). |
| Multiply through | e^(2x)y′ + 2e^(2x)y = 6e^(2x), or d(e^(2x)y)/dx = 6e^(2x). |
| Integrate and divide | e^(2x)y = 3e^(2x) + C, so y = 3 + Ce^(−2x). |
| Initial condition | 1 = 3 + C gives C = −2; y = 3 − 2e^(−2x). |
Check your method
The solution tends to the equilibrium y = 3 as x increases. Substitution gives y′ + 2y = 4e^(−2x) + 6 − 4e^(−2x) = 6. Do not use the original coefficient of y as P until the coefficient of y′ has been reduced to 1.
Try it yourself
Solve 2y′ + 4y = 8 with y(0) = 0.
Show the answer and method
First divide by 2: y′ + 2y = 4. The general solution is y = 2 + Ce^(−2x), and C = −2. Therefore y = 2 − 2e^(−2x).
This written practice example was prepared for Jemshah revision. It is additional practice, not an official examination question or a transcript of the video.