Find the general solution of y(1-ln y) y'' + (1+sin y)…

Find the general solution of $y(1-\ln y)\,y'' + (1+\sin y)\,(y')^2 = 0$.

Step 1: Substitution

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

The equation becomes: \[y(1-\ln y)\,p\,\frac{dp}{dy} + (1+\sin y)\,p^2 = 0. \tag{1}\]

Step 2: Separate variables

Step 3: Integrate the right-hand side

Split the fraction: \[\frac{1+\sin y}{y(1-\ln y)} = \frac{1}{y(1-\ln y)} + \frac{\sin y}{y(1-\ln y)}.\]

For the first part, use $u = 1-\ln y$, $du = -\dfrac{dy}{y}$: $\int \frac{dy}{y(1-\ln y)} = -\int \frac{du}{u} = -\ln|1-\ln y|.$

The second part, $\displaystyle\int\frac{\sin y}{y(1-\ln y)}\,dy$, has no closed elementary form in general; denote it

\[ F(y) = \int\frac{\sin y}{y(1-\ln y)}\,dy. \]

Step 4: Integrate equation (2)

\[ \ln|p| = \ln|1-\ln y| - F(y) + \ln|C_1|. \]

Thus: \[p = C_1(1-\ln y)\,e^{-F(y)}. \tag{3}\]

Step 5: General solution

Since $p = y' = dx/dy^{-1}$, a further quadrature $dx = dy/p$ gives $x$ as a function of $y$:

Step 6: Verification

Differentiate $p = C_1(1-\ln y)e^{-F(y)}$ with respect to $y$, using $F'(y) = \dfrac{\sin y}{y(1-\ln y)}$: \[\frac{dp}{dy} = -C_1 e^{-F(y)}\left[\frac{1}{y} + (1-\ln y)\cdot\frac{\sin y}{y(1-\ln y)}\right] = -C_1 e^{-F(y)}\cdot\frac{1+\sin y}{y}.\]

Then: $y'' = p\frac{dp}{dy} = -\frac{C_1^2(1-\ln y)(1+\sin y)}{y}e^{-2F(y)}.$

Check $y(1-\ln y)y'' + (1+\sin y)(y')^2$: $-C_1^2(1-\ln y)^2(1+\sin y)e^{-2F(y)} + C_1^2(1-\ln y)^2(1+\sin y)e^{-2F(y)} = 0.$ $\checkmark$