Question

The differential equation whose solution is $y - Ae^{3x} + Be^{-3x}$ is given by

• A. $y_2 - 3y_1 + 3y = 0$
• B. $xy_{2}+3y_{1}-xy+x^{2}+3=0$
• C. $y_2-9y=0$
• D. $\left(y_{1}\right)^{3}-3y\left(xy_{2}-3y\right)=0$

Answer 1

Answer: (C)

We have, $y = Ae^{3x} + Be^{-3x}$
Differentiating w.r.t. $x$, we get
$y_1 = 3Ae^{3x} -3\,Be^{-3x}$
Again differentiating w.r.t. $x$, we get
$y_{2}=9\,Ae^{3x}+9\,Be^{-3x}=9\left(Ae^{3x}+Be^{-3x}\right)=9y$
$\Rightarrow y_{2}-9y=0$