Question

The solution of the differential equation $y^{\prime \prime \prime }-8y^{\prime \prime }=0,$ where $y\left(0\right)=\frac{1}{8},y^{\prime }\left(0\right)=0,y^{\prime \prime }\left(0\right)=1$ is:

• A. $y=\frac{1}{8}(\frac{e^{8x}}{8}+x-\frac{7}{9})$
• B. $y=\frac{1}{8}(\frac{e^{8x}}{8}+x+\frac{7}{9})$
• C. $y=\frac{1}{8}(\frac{e^{8x}}{8}-x+\frac{7}{8})$
• D. none of these

Answer 1

Answer: (C)

$y=\frac{1}{8}(\frac{e^{8x}}{8}-x+\frac{7}{8})$

$y^{\prime \prime \prime }-8y^{\prime \prime }=0$
$\Rightarrow \frac{y^{\prime \prime \prime }}{y^{\prime \prime }}=8$
Integrating we get, $\Rightarrow \ln y^{\prime \prime }=8x+c$
given, $y^{\prime \prime }\left(0\right)=1\Rightarrow \log 1=0+c\therefore c=0$
$y^{\prime \prime }=e^{8x}$
Integrating we get, $y^{\prime }=\frac{e^{8x}}{8}+d$
Also given $y^{\prime }\left(0\right)=0,d=-\frac{1}{8}$
$y^{\prime }=\frac{e^{8x}}{8}-\frac{1}{8}$
Again integrating, $\Rightarrow y=\frac{e^{8x}}{64}-\frac{x}{8}+k$
also $y\left(0\right)=\frac{1}{8}\therefore k=\frac{7}{64}\therefore y=\frac{1}{8}(\frac{e^{8x}}{8}-x+\frac{7}{8})$, which is required solution.