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

$y=e^x(a\;\cos x+b\;\sin x)$ $\:$:$\:$ $\frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=0$
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}= $$0$. The product rule of differentiation states $uv = uv' + vu'$
Step 1:
Given :$ y = e^x.a\cos x + e^xb\sin x$
Differentiating on both sides with respect to $x$ we get,
$\large\frac{dy}{dx} =$$ e^x.(-a\sin x) + e^x.a\cos x + e^x.b\cos x + e^x.b\sin x$
$\quad\;\;= (a+b)e^x\cos x + (b - a)e^x\sin x$
Differentiating again with respect to $x$ we get,
$\Rightarrow (a+b)[e^x(-\sin x)+ e^x.\cos x] + (b - a)[e^x\cos x + e^x\sin x]$
$\large\frac{d^2y}{dx^2 }$$= - ae^x\sin x + ae^x \cos x - be^x \sin x +be^x\cos x + be^x\cos x +be^x\sin x - ae^x\cos x -ae^x\sin x$
On simplifying we get,
$2e^x[b\cos x - a\sin x]$
Step 2:
substituting for $\large\frac{d^2y}{dx^2} ,\frac{dy}{dx}$ and $y$ in the solution we get,
$2e^x[b\cos x - a\sin x] - {2[(a+b)e^x\cos x + (b - a) e^x\sin x ]}+2e^x[a\cos x + b\sin x]$
$2be^x\cos x -2ae^x\sin x -2ae^x\cos x -2be^x\cos x - 2be^x\sin x + 2ae^x\sin x + 2ae^x\cos x + 2be^x\sin x = 0$
Hence the given function is a solution to 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