Solving the ODE : $y' = \ln(x^2 + 2x + y)$

Question

$$y'= \ln(x²+2x+y)$$ Can you use laplace transform to solve this equation, $y=f(x)$ and $y'=f'(x)$ Wolfram seems to take too much time. i tried substituting $u=y+x²+2x$ which gave $u'-2x-2=\ln(u)$ i was stuck from there

Answer 1

I do not think there is analytical solution. But you can try the next best thing, which is series solution?

\begin{gather*} \boxed{y^{\prime}-\ln \left(x^{2}+2 x +y\right)=0} \end{gather*} With the expansion point for the power series method at $x = 0$. Using series expansion we have \begin{equation} y\left( x\right) =y_{0}+\sum_{n=0}^{\infty}\frac{x^{n+1}}{\left( n+1\right) !}\left. F_{n}\right\vert _{x_{0},y_{0}} \tag{1} \end{equation} Where \begin{align*} F_0 &= \ln \left(x^{2}+2 x +y\right)\\ F_1 &= \frac{d F_0}{dx} \\ &= \frac{\partial F_0}{\partial x}+ \frac{\partial F_0}{\partial y} F_0 \\ &= \frac{2 x +2+\ln \left(x^{2}+2 x +y\right)}{x^{2}+2 x +y}\\ F_2 &= \frac{d F_1}{d ,x} \\ &= \frac{\partial F_1}{\partial x}+ \frac{\partial F_1}{\partial y} F_1 \\ &= \frac{-\ln \left(x^{2}+2 x +y\right)^{2}+\left(-4 x -3\right) \ln \left(x^{2}+2 x +y\right)-2 x^{2}-2 x +2 y-2}{\left(x^{2}+2 x +y\right)^{2}} \end{align*} And so on. Evaluating all the above at initial conditions $x \left(0\right) = 0$ and $y \left(0\right) = y \left(0\right)$ gives \begin{align*} F_0 &= \ln \left(y \left(0\right)\right)\\ F_1 &= \frac{2+\ln \left(y \left(0\right)\right)}{y \left(0\right)}\\ F_2 &= \frac{-\ln \left(y \left(0\right)\right)^{2}-3 \ln \left(y \left(0\right)\right)-2+2 y \left(0\right)}{y \left(0\right)^{2}} \end{align*} Substituting all the above in (1) and simplifying gives the solution as \begin{align*} y = y \left(0\right)+\ln \left(y \left(0\right)\right) x +\frac{\frac{x^{2} \ln \left(y \left(0\right)\right)}{2}+x^{2}+\frac{x^{3}}{3}}{y \left(0\right)}+\frac{-\frac{x^{3} \ln \left(y \left(0\right)\right)^{2}}{6}-\frac{x^{3} \ln \left(y \left(0\right)\right)}{2}-\frac{x^{3}}{3}}{y \left(0\right)^{2}}+O\left(x^{4}\right) \end{align*}