Question

Find the solution of $\frac{d^{2}y}{d{x}^2}=y$ which passes through origin and point $(ln2,\frac{3}{4})$

• A. $y=\frac{1}{2}{e}^x-e^{-x}$
• B. $y=\frac{1}{2}(e^x+e^{-x})$
• C. $y=\frac{1}{2}(e^x-e^{-x})$
• D. $y=\frac{1}{2}{e}^x+e^{-x}$

Answer 1

The correct option is C $y=\frac{1}{2}(e^x-e^{-x})$

$\frac{d^{2}y}{d{x}^2}=y$
Multiply $2\left(\frac{dy}{dx}\right)$ both sides,
$2\frac{dy}{dx}\frac{d^{2}y}{d{x}^x}=2y\frac{dy}{dx}$
$\frac{d}{dx}{\left(\frac{dy}{dx}\right)}^2=\frac{d}{dx}\left(y^2\right)$
Integrating both sides,
${\left(\frac{dy}{dx}\right)}^2=y^2+C_1^2$
$\frac{dy}{dx}=\sqrt{y^2+C_1^2}$
$\int \frac{dy}{\sqrt{y^2+C_1^2}}=\int dx$
$ln[y+\sqrt{y^2+C_1^2}]=x+C_2$
It passes through $(0,0)$
$\Rightarrow$ log$\left[C_1\right]$ = $\left[C_2\right]$
So, $ln[y+\sqrt{y^2+C_1^2}]=x+log{C}_1$
$\Rightarrow$ $ln\left[\frac{y+\sqrt{y^2+C_1^2}}{C_1}\right]=x...$ (i)
It passes through $(ln2,3/4)$
$ln\left[\frac{\frac{3}{4}+\sqrt{\frac{9}{16}+C_1^2}}{C_1}\right]=ln2$
$\frac{3}{4}+\sqrt{\frac{9}{16}+C_1^2}=2C_1$
$\sqrt{C_1^2+\frac{9}{16}}=2C_1-3/4$
Square both sides
$C_1^2+\frac{9}{16}=4C_1^2+\frac{9}{16}-3C_1$
$3C_1^2-3C_1=0$
$C_1=0,1$
$C_1=0$ is not possible
$C_1=1$
Hence, by (i)
$y+\sqrt{y^2+1}=e^x$
$\sqrt{y^2+1}=e^x-y$
Square both sides,
$y^2+1=e^{2x}+y^2-2e^{x}y$
$1=e^{2x}-2e^{x}y$
$2e^{x}y=e^{2x}-1$
$y=\frac{1}{2}(e^x-e^{-x})$
Alternative Solution:
By options we can obtain that the curve $y=\frac{1}{2}(e^x-e^{-x})$ is passing through $(0,0)$ & $(ln2,3/4)$