Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Differential Equations
0 votes

Verify that the given function (implicit or explicit) is a solution of the corresponding differential equation

$y=x\;\sin 3x$ $\:$:$\:$ $\frac{d^2y}{dx^2}+9y-6\;\cos 3x=0$
Can you answer this question?

1 Answer

0 votes
  • $\large\frac{d(e^x)}{dx }$$= e^x$; $\large\frac{d(x)}{dx}$$ = 1$; $\large\frac{d(constant\; term )} {dx}= $$0$. The product rule of differentiation states $uv = uv' + vu'$
Step 1:
Given $y = x\sin 3x$
Differentiating on both sides we get,
$\large\frac{dy}{dx} $$= x. (3\cos3x) + (\sin3x).1$
$\large\frac{dy}{dx} $$= 3x\cos 3x + \sin 3x$
differentiating again we get
$\large\frac{d^2y}{dx^2 }$$= 3x.(- 3\sin 3x) + 3x(\cos 3x) + (3\cos 3x)$
$\large\frac{d^2y}{dx^2} $$= -9x\sin 3x + 3x\cos 3x +3\cos 3x$
Step 2:
Substituting in the given solution for $\large\frac{d^2y}{dx^2},\large\frac{dy}{dx}$ and $y$ we get
$-9x\sin 3x + 3x\cos 3x + 3\cos 3x + 9(x\sin 3x) - 6\cos 3x =0$
Hence the given function is the solution for the given differential equation.
answered Jul 30, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App