Recognise and solve a separable first-order equation
An ordinary differential equation relates an unknown function to derivatives with respect to one independent variable. Its order is the highest derivative present. A separable first-order equation can be rearranged so that all terms in the dependent variable are integrated on one side and the independent variable on the other. When dividing by a function of y, check separately whether the excluded values give constant solutions.
Worked practice example
Solve dy/dx = 3x²y with y(0) = 2.
| Step | Working |
|---|---|
| Separate | For y ≠ 0, dy/y = 3x² dx. |
| Integrate | ln|y| = x³ + C, so y = Ae^(x³). |
| Use the initial value | At x = 0, 2 = A; hence y = 2e^(x³). |
| Verify | y′ = 6x²e^(x³) = 3x²y, and y(0) = 2. |
Check your method
The zero function is also a solution of the differential equation, although it does not meet this initial condition. One initial condition fixes the one arbitrary constant in this first-order problem. Differentiate the final expression rather than checking only the initial value.
Try it yourself
Solve dy/dx = −2y with y(0) = 5.
Show the answer and method
Separate and integrate to obtain y = Ae^(−2x). The initial condition gives A = 5, so y = 5e^(−2x). The negative coefficient produces decay as x increases.
This written practice example was prepared for Jemshah revision. It is additional practice, not an official examination question or a transcript of the video.