Find the differential equation whose general solution is y =…

Find the differential equation whose general solution is $y = c_1\sin x + c_2 x$, where $c_1$ and $c_2$ are arbitrary constants.

Step 1: Differentiate the general solution

Differentiating $y = c_1\sin x + c_2 x$ twice gives

\[y' = c_1\cos x + c_2, \qquad (1)\]

\[y'' = -c_1\sin x. \qquad (2)\]

Step 2: Eliminate the arbitrary constants

From (2), $c_1 = -\dfrac{y''}{\sin x}$. Substituting into the original equation, $y = -\frac{y''}{\sin x}\cdot\sin x + c_2 x = -y'' + c_2 x$, so \[c_2 = \frac{y + y''}{x}. \qquad (4)\]

Substitute $c_1$ into (1): $y' = -y''\cot x + c_2.$

Then use (4): $y' = -y''\cot x + \frac{y + y''}{x}.$

Step 3: Rearrange into standard form

Multiply through by $x$ and rearrange: $y''(1 - x\cot x) - xy' + y = 0.$

Step 4: Verification

With $y = c_1\sin x + c_2 x$, $y' = c_1\cos x + c_2$, and $y'' = -c_1\sin x$, \[y''(1-x\cot x) - xy' + y = -c_1\sin x\left(1 - x\frac{\cos x}{\sin x}\right) - x(c_1\cos x + c_2) + c_1\sin x + c_2 x\]

\[= -c_1\sin x + c_1 x\cos x - c_1 x\cos x - c_2 x + c_1\sin x + c_2 x = 0.\]

The restriction $x \neq k\pi$ ($k\in\mathbb{Z}$) is needed so that $\cot x$ is defined and $\sin x\neq 0$, which makes the elimination of $c_1$ valid.