Find the general solution of y''sin y = 2(y')^2cos y

Find the general solution of $y''\sin y = 2(y')^2\cos y$.

Step 1: Substitution

Let $p = y'$. Express $y''$ using the chain rule: $y'' = p\,\frac{dp}{dy}.$

The equation becomes: \[p\,\frac{dp}{dy}\sin y = 2p^2\cos y. \tag{1}\]

Step 2: Separate variables

Step 3: Integrate both sides

\[ \int\frac{dp}{p} = 2\int\cot y\,dy,\qquad \ln|p| = 2\ln|\sin y| + \ln|C_1|,\qquad p = C_1\sin^2 y. \]

Thus: \[y' = C_1\sin^2 y. \tag{2}\]

Step 4: Second integration

Separate variables in (2): \[\frac{dy}{\sin^2 y} = C_1\,dx,\qquad \int\csc^2 y\,dy = \int C_1\,dx,\qquad -\cot y = C_1 x + C_2.\]

Step 5: General solution

or equivalently $\cot y = -C_1\,x - C_2$, where $C_1$ and $C_2$ are arbitrary constants, valid for $\sin y \neq 0$.

Step 6: Verification

From $-\cot y = C_1 x + C_2$, differentiate implicitly: $\csc^2 y\cdot y' = C_1,\qquad y' = C_1\sin^2 y.$

Differentiate again: $y'' = C_1\cdot 2\sin y\cos y\cdot y' = 2C_1^2\sin^3 y\cos y.$

Check $y''\sin y - 2(y')^2\cos y$: \[2C_1^2\sin^3 y\cos y\cdot\sin y - 2C_1^2\sin^4 y\cdot\cos y = 2C_1^2\sin^4 y\cos y - 2C_1^2\sin^4 y\cos y = 0.\ \checkmark\]