Question

solve the given differential equation. (x^{2}+9)dy=(2x - 8xy)dx

Explanation:

Step1: Rearrange the differential equation

Separate the variables: $\frac{dy}{2x - 8xy}=\frac{dx}{x^{2}+9}$. Factor out $2x$ from the denominator on the left - hand side: $\frac{dy}{2x(1 - 4y)}=\frac{dx}{x^{2}+9}$.

Step2: Integrate both sides

Integrate $\int\frac{dy}{2x(1 - 4y)}=\int\frac{dx}{x^{2}+9}$. First, integrate the left - hand side with respect to $y$ and the right - hand side with respect to $x$. For $\int\frac{dy}{1 - 4y}$, let $u = 1-4y$, then $du=-4dy$ and $\int\frac{dy}{1 - 4y}=-\frac{1}{4}\ln|1 - 4y|$. So, $\int\frac{dy}{2x(1 - 4y)}=\frac{1}{2x}\int\frac{dy}{1 - 4y}=-\frac{1}{8x}\ln|1 - 4y|+C_1$. For $\int\frac{dx}{x^{2}+9}$, since $\int\frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\tan^{- 1}(\frac{x}{a})+C$ with $a = 3$, we have $\int\frac{dx}{x^{2}+9}=\frac{1}{3}\tan^{-1}(\frac{x}{3})+C_2$.

Step3: Solve for $y$

We have $-\frac{1}{8x}\ln|1 - 4y|=\frac{1}{3}\tan^{-1}(\frac{x}{3})+C$. Multiply both sides by $- 8x$: $\ln|1 - 4y|=- \frac{8x}{3}\tan^{-1}(\frac{x}{3})-8Cx$. Exponentiate both sides: $|1 - 4y|=e^{-\frac{8x}{3}\tan^{-1}(\frac{x}{3})-8Cx}$. Then $1 - 4y = \pm e^{-8Cx}e^{-\frac{8x}{3}\tan^{-1}(\frac{x}{3})}$. Let $k=\pm e^{-8C}$, so $y=\frac{1 - ke^{-\frac{8x}{3}\tan^{-1}(\frac{x}{3})}}{4}$.

Answer:

$y=\frac{1 - ke^{-\frac{8x}{3}\tan^{-1}(\frac{x}{3})}}{4}$, where $k$ is an arbitrary constant.