Question

find the general solution using separable method.
a. $\frac{d^{2}y}{dx^{2}} = xsin x$
b. $\frac{dy}{dx}=\frac{1 - 2xy}{1 + x^{3}}$
c. $(x+\frac{1}{x})\frac{dy}{dx}+y = 0$

Explanation:

Step1: Solve part A

We have $\frac{d^{2}y}{dx^{2}}=x\sin x$. First, integrate both sides with respect to $x$ to get $\frac{dy}{dx}$. Using integration - by - parts $\int u dv=uv-\int v du$, let $u = x$, $dv=\sin xdx$, then $du = dx$, $v=-\cos x$. So $\int x\sin xdx=-x\cos x+\int\cos xdx=-x\cos x+\sin x + C_1$. Then integrate $\frac{dy}{dx}=-x\cos x+\sin x + C_1$ with respect to $x$ again. $\int(-x\cos x+\sin x + C_1)dx$. For $\int - x\cos xdx$, use integration - by - parts again. Let $u=-x$, $dv = \cos xdx$, then $du=-dx$, $v=\sin x$, so $\int - x\cos xdx=-x\sin x-\int(-\sin x)dx=-x\sin x-\cos x$. And $\int\sin xdx=-\cos x$, $\int C_1dx = C_1x$. So $y=-x\sin x-\cos x-\cos x + C_1x + C_2=-x\sin x-2\cos x + C_1x + C_2$.

Step2: Solve part B

We have $\frac{dy}{dx}=\frac{1 - 2xy}{1 + x^{3}}$. Rearrange it to the form of a first - order linear differential equation $\frac{dy}{dx}+\frac{2x}{1 + x^{3}}y=\frac{1}{1 + x^{3}}$. The integrating factor is $\mu(x)=e^{\int\frac{2x}{1 + x^{3}}dx}$. Let $u = 1 + x^{3}$, then $du = 3x^{2}dx$, and $\int\frac{2x}{1 + x^{3}}dx=\frac{2}{3}\int\frac{1}{u}du=\frac{2}{3}\ln|1 + x^{3}|$, so $\mu(x)=(1 + x^{3})^{\frac{2}{3}}$. Multiply the entire equation by the integrating factor: $(1 + x^{3})^{\frac{2}{3}}\frac{dy}{dx}+2x(1 + x^{3})^{-\frac{1}{3}}y=(1 + x^{3})^{-\frac{1}{3}}$. The left - hand side is the derivative of $y(1 + x^{3})^{\frac{2}{3}}$. Integrating both sides gives $y(1 + x^{3})^{\frac{2}{3}}=\int(1 + x^{3})^{-\frac{1}{3}}dx + C$. But $\int(1 + x^{3})^{-\frac{1}{3}}dx$ cannot be expressed in terms of elementary functions.

Step3: Solve part C

We have $(x+\frac{1}{x})\frac{dy}{dx}+y = 0$. First, rewrite it as $(x+\frac{1}{x})\frac{dy}{dx}=-y$. Then separate the variables: $\frac{dy}{y}=-\frac{dx}{x+\frac{1}{x}}=-\frac{x dx}{x^{2}+1}$. Integrate both sides. $\int\frac{dy}{y}=-\frac{1}{2}\int\frac{2x dx}{x^{2}+1}$. Let $u = x^{2}+1$, then $du = 2x dx$. So $\ln|y|=-\frac{1}{2}\ln|x^{2}+1|+C=\ln\frac{1}{\sqrt{x^{2}+1}}+C$. Then $y=\frac{C}{\sqrt{x^{2}+1}}$.

Answer:

A. $y=-x\sin x-2\cos x + C_1x + C_2$
B. $y(1 + x^{3})^{\frac{2}{3}}=\int(1 + x^{3})^{-\frac{1}{3}}dx + C$ (where $\int(1 + x^{3})^{-\frac{1}{3}}dx$ is a non - elementary integral)
C. $y=\frac{C}{\sqrt{x^{2}+1}}$