Search any question & find its solution
Question:
Answered & Verified by Expert
If $\mathrm{c}$ and $\mathrm{d}$ are arbitrary constants, then $y=e^{2 x}(\cosh \sqrt{2} x+d \sinh \sqrt{2} x)$ is the general solution of the differential equation
Options:
Solution:
1181 Upvotes
Verified Answer
The correct answer is:
$y^{\prime \prime}-4 y^{\prime}+2 y=0$
$y=e^{2 x}(c \cos h \sqrt{2} x+d \sin h \sqrt{2} x)$
Then, $y^{\prime}=e^{2 x}(2 c \cos h \sqrt{2} x+2 d \sin h \sqrt{2} x$
$+\sqrt{2} c \sin h \sqrt{2} x d \sqrt{2} \cos h \sqrt{2} x)$
$\begin{array}{r}
\Rightarrow y^{\prime}=e^{2 x}[\cos h \sqrt{2} x(2 c+d \sqrt{2})+2 \sin h \sqrt{2} x(2 d+\sqrt{2} c)] \\
\Rightarrow y^{\prime \prime}=e^{2 x}[2 \cos h \sqrt{2} x(2 c+d \sqrt{2})+2 \sin h \sqrt{2} x \\
(2 d+\sqrt{2} c)+\sqrt{2}(2 c+d \sqrt{2}) \sin h \sqrt{2} x \\
+\sqrt{2}(2 d+\sqrt{2} c) \cos h \sqrt{2} x]
\end{array}$
Now, $y^{\prime \prime}-4 y^{\prime}+2 y=0$
Then, $y^{\prime}=e^{2 x}(2 c \cos h \sqrt{2} x+2 d \sin h \sqrt{2} x$
$+\sqrt{2} c \sin h \sqrt{2} x d \sqrt{2} \cos h \sqrt{2} x)$
$\begin{array}{r}
\Rightarrow y^{\prime}=e^{2 x}[\cos h \sqrt{2} x(2 c+d \sqrt{2})+2 \sin h \sqrt{2} x(2 d+\sqrt{2} c)] \\
\Rightarrow y^{\prime \prime}=e^{2 x}[2 \cos h \sqrt{2} x(2 c+d \sqrt{2})+2 \sin h \sqrt{2} x \\
(2 d+\sqrt{2} c)+\sqrt{2}(2 c+d \sqrt{2}) \sin h \sqrt{2} x \\
+\sqrt{2}(2 d+\sqrt{2} c) \cos h \sqrt{2} x]
\end{array}$
Now, $y^{\prime \prime}-4 y^{\prime}+2 y=0$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.