Final answer:
To find the general solution of the given differential equation y' = 2yx² + 7, we can separate the variables and integrate both sides of the equation. The general solution is y = sqrt(e^(2x² + 2C) - 7) or y = -sqrt(e^(2x² + 2C) - 7).
Step-by-step explanation:
The given differential equation is y' = 2yx² + 7.
To find the general solution, we can separate the variables and integrate both sides of the equation. Here are the steps:
- Divide both sides of the equation by y² + 7: y' / (y² + 7) = 2x.
- Integrate both sides with respect to y and x: ∫(1 / (y² + 7)) dy = ∫2x dx.
- Integrate the left side using the substitution u = y² + 7, which gives: (1 / 2) ln|y² + 7| = x² + C.
- Exponentiate both sides of the equation to get rid of the natural logarithm: ln|y² + 7| = 2x² + 2C.
- Remove the logarithm by taking the exponent of both sides: |y² + 7| = e^(2x² + 2C).
- Since the absolute value of y² + 7 can be positive or negative, we have two cases: y² + 7 = e^(2x² + 2C) or y² + 7 = -e^(2x² + 2C).
- Finally, solve each case for y to obtain the general solution.
Therefore, the general solution of the given differential equation is y = sqrt(e^(2x² + 2C) - 7) or y = -sqrt(e^(2x² + 2C) - 7).