Question #3829a

Jul 4, 2017

$y \left(x\right) = 4 \sqrt{\frac{1 + x}{3 + x}}$

Explanation:

Perform partial fractions decomposition

$\frac{1}{\left(1 + x\right) \left(3 + x\right)} = \frac{A}{1 + x} + \frac{B}{3 + x}$

$\frac{1}{\left(1 + x\right) \left(3 + x\right)} = \frac{A \left(3 + x\right) + B \left(1 + x\right)}{\left(1 + x\right) \left(3 + x\right)}$

$\frac{1}{\left(1 + x\right) \left(3 + x\right)} = \frac{\left(A + B\right) x + \left(3 A + B\right)}{\left(1 + x\right) \left(3 + x\right)}$

$\left\{\begin{matrix}A + B = 0 \\ 3 A + B = 1\end{matrix}\right.$

$\left\{\begin{matrix}A = - B \\ - 3 B + B = 1\end{matrix}\right.$

$\left\{\begin{matrix}A = \frac{1}{2} \\ B = - \frac{1}{2}\end{matrix}\right.$

So:

$\frac{1}{\left(1 + x\right) \left(3 + x\right)} = \frac{1}{2 \left(1 + x\right)} - \frac{1}{2 \left(3 + x\right)}$

Solve now:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{\left(1 + x\right) \left(3 + x\right)}$

Separate the variables:

$\frac{\mathrm{dy}}{y} = \frac{\mathrm{dx}}{\left(1 + x\right) \left(3 + x\right)}$

$\ln \left\mid y \right\mid = \int \frac{\mathrm{dx}}{\left(1 + x\right) \left(3 + x\right)} = \frac{1}{2} \int \frac{\mathrm{dx}}{1 + x} - \frac{1}{2} \int \frac{\mathrm{dx}}{3 + x} + C$

$\ln \left\mid y \right\mid = \frac{1}{2} \ln \left\mid 1 + x \right\mid - \frac{1}{2} \ln \left\mid 3 + x \right\mid + C$

For $x > - 1$ both arguments are positive, so:

$\ln \left\mid y \right\mid = \frac{1}{2} \ln \left(1 + x\right) - \frac{1}{2} \ln \left(3 + x\right) = \ln \sqrt{\frac{1 + x}{3 + x}} + C$

Taking the exponential:

$\left\mid y \right\mid = {c}_{1} \sqrt{\frac{1 + x}{3 + x}}$

where ${c}_{1} = {e}^{C}$ is positive. Admitting also negative value for ${c}_{1}$ and noting that $y \left(x\right) = 0$ is also a solution, then the general solution is:

$y \left(x\right) = c \sqrt{\frac{1 + x}{3 + x}}$ with $c \in \mathbb{R}$

We can determine the particular solution for:

$y \left(1\right) = 2$

$c \sqrt{\frac{1 + 1}{3 + 1}} = 2$

$c \sqrt{\frac{1}{4}} = 2$

$\frac{c}{2} = 2$

$c = 4$

Finally:

$y \left(x\right) = 4 \sqrt{\frac{1 + x}{3 + x}}$