I want to ask if a differential equation of second order can be solved by integration? Like equations of the type $\dfrac{d^2y}{dx^2} = f(y)$. I know this can be solved by making equations of the form $(D² - a²)y = c$ Where $D$ is the operator $\dfrac{d(•)}{dx}$. But I want to know can this be solved using direct integration. In particular my equation is $\dfrac{d^2y}{dx^2} = (1-y)$.
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$\begingroup$Indeed, it can. Illustrating with your equation: we have $$ ((y')^2)'=2y'y''=2y'(1-y) $$ and so $$ ((y')^2)'=(2y-y^2)'. $$ Integrating on both sides gives $$ (y')^2=2y-y^2+c $$ for some $c$ and so $$ y'=\pm\sqrt{2y-y^2+c}. $$ You need to solve the last equation, for each $c$ and each sign $\pm$, and then substitute in the original equation. This last step is necessary in view of the $\pm$ signs since you don't know if you got more than the solutions of the original equation.
$\endgroup$ 1 $\begingroup$Yes, it can be solved by direct integration. Let $u=y'+iy$ then from $y''+y=1$ we obtain $$e^{-ix}(y''+y)=e^{-ix}(u'-iu)=D(e^{-ix}u)=e^{-ix}\cdot 1$$ By integrating it, we get $$e^{-ix}u=ie^{-ix}+c_1$$ that is, after multiplying both sides by $e^{2ix}$, $$D(e^{ix}y)=e^{ix}(y'+iy)=e^{ix}i+c_1e^{2ix}.$$ Again, by integrating it, we get $$e^{ix}y=e^{ix}-2ic_1e^{2ix}+c_2$$ that is $$y=1-2ic_1e^{ix}+c_2e^{-ix}=1+C_1e^{ix}+C_2e^{-ix}$$ where $C_1$ and $C_2$ are arbitrary complex constants.
$\endgroup$ 2 $\begingroup$In general for a equation of the form $f(y,y^\prime,y^{\prime\prime})=0$ which $x$ doesn't appear in it, we let $y^\prime=u$ then $$y^{\prime\prime}=\frac{du}{dx}=\frac{du}{dy}\frac{dy}{dx}=uu^\prime$$ and the equation will be of the form $f(y,u,uu^\prime)=0$. This is a equation of the first order of $u$ respect to $y$, where $u^\prime=\dfrac{du}{dy}$.
For your example $y''=1-y$, then $uu'=1-y$ or $u\dfrac{du}{dy}=1-y$ which can be solved by separation.
$\endgroup$ 1 $\begingroup$The general case is $$\frac{d^{2}y(x)}{dx^{2}}=Y(y(x))$$ One has $$\frac{d^{2}y(x)}{dx^{2}}=\frac{d}{dx}\frac{dy(x)}{dx}=\frac{dy}{dx}\frac{d}{dy}\frac{dy(x)}{dx}=y'\frac{dy'}{dy}$$ Thus $$y'\frac{dy'}{dy}=Y(y)$$ Or $$y'dy'=Y(y)dy$$ Integrating the both sides we get $$(y')^{2}=2\int^{y}Y(z)dz+C_{1}$$ Or, $$y'(x)=\pm\sqrt{2\int^{y}Y(z)dz+C_{1}}$$ The equation is separble, thus $$\pm(x+C_{2})=\int^{y}\frac{dw}{\sqrt{2\int^{w}Y(z)dz+C_{1}}}$$
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