Question

The curve satisfying $\frac{d^{2}y}{d{x}^2}-4y^{\prime }=0$ and passing through $(1,0)$ is:

• A. $y=x{e}^{4x}$
• B. $y=a{e}^{4x}$
• C. $y=e^{4x}-e^4$
• D. $y=a(e^{4x}-e^4)$

Answer 1

The correct option is D $y=a(e^{4x}-e^4)$

We have, $y^{\prime \prime }-4y^{\prime }=0$ .....(1)
Put $z=y^{\prime }\Rightarrow z^{\prime }=y^{\prime \prime }$
Thus (1) becomes, $z^{\prime }-4z=0$
$\Rightarrow \frac{dz}{z}=4dx$
Integrating we get,
$\ln z=4x+c$
$\Rightarrow z=e^{4x+c}$
$\Rightarrow \frac{dy}{dx}=e^{4x+c}$
Integrating we get, $y=\frac{e^{4x+c}}{4}+k$
Now using $y\left(1\right)=0\Rightarrow 0=\frac{e^{4+c}}{4}+k\Rightarrow k=-\frac{e^{4+c}}{4}$
Thus solution is,
$y=\frac{e^{4x+c}-e^{4+c}}{4}=\frac{e^c}{4}(e^{4x}-e^4)=a(e^{4x}-e^4)$
Where $a=\frac{e^c}{4}=$ any constant