Find the general solution of the differential equation. Solve for as a function of .

Question

Find the general solution of the differential equation.  Solve for  as a function of  .

Quizplus · Accepted Answer

The given differential equation can be written as

$\frac{dy}{dx}=\sqrt{1+y^2}-y$

Separating the variables, we get

$\int \frac{1}{\sqrt{1+y^2}}dy = \int 1 dx$

$\ln \bigg|y+\sqrt{1+y^2}\bigg|=x+C$

Simplifying this equation, we get

$y+\sqrt{1+y^2}=\pm e^{x+C}$

Squaring both sides and rearranging the terms, we get

$y^2=e^{2x+2C}-1$

Therefore,

$y=\pm\sqrt{e^{2x+2C}-1}$

Now, using the initial condition $y(0)=2$, we get

$2=\pm\sqrt{e^{2\cdot0+2C}-1}$

$4=e^{2C}-1$

$e^{2C}=5$

Thus,

$y=\pm\sqrt{5e^{2x}-5}$

Simplifying further, we get

$y=\pm\sqrt{5}(e^x)^{\frac{3}{2}}\sqrt{e^{-x}-\frac{1}{5}}$

Hence,

$y=\pm\sqrt{5}x^{\frac{3}{2}}\sqrt{1-\frac{1}{5x}+\frac{1}{25x^2}-\frac{16}{375x^3}+...}$

Therefore,

$y=\pm\sqrt{5}x^{\frac{3}{2}}-\frac{\sqrt{5}}{2}+\mathcal O\bigg(\frac{1}{x^{\frac{1}{2}}}\bigg)$

Thus,

$y\sim \pm\sqrt{5}x^{\frac{3}{2}}$ as $x	o\infty$

Find the General Solution of the Differential Equation