Problem 6
Solve the differential equation y'' = x + sin x by…
Solve the differential equation $y'' = x + \sin x$ by successive integration.
Explanation of Method
The right-hand side depends only on $x$, so integrate twice, introducing a new arbitrary constant at each step.
Step 1: First integration
\[y' = \int(x + \sin x)\,dx = \frac{x^2}{2} - \cos x + C_1.\]
Step 2: Second integration
\[y = \int\left(\frac{x^2}{2} - \cos x + C_1\right)dx = \frac{x^3}{6} - \sin x + C_1 x + C_2.\]
Step 3: General solution
\[y = \frac{x^3}{6} - \sin x + C_1 x + C_2,\] where $C_1$ and $C_2$ are arbitrary constants.
Step 4: Verification
Differentiating twice: $y' = \frac{x^2}{2} - \cos x + C_1$, $y'' = x + \sin x$.
Result
Final answer:
\[\boxed{\,y = \dfrac{x^3}{6} - \sin x + C_1 x + C_2\,}\]