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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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verify that the given function (implicit or explicit) is a solution of the corresponding differential equation

function: $y=ae^x+be^{-x}+x^2$
diff. eqn., $x\frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy + x^2 -2 = 0$
Can you answer this question?
 
 

1 Answer

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Toolbox:
  • $\large\frac{d(e^x)}{dx }$$= e^x$; $\large\frac{d(x)}{dx}$$ = 1$; $\large\frac{d(constant\; term )} {dx}= $$0$. The product rule of differentiation states $uv = uv' + vu'$
Step 1:
Given :$y=ae^x+be^{-x}+x^2$
Differentiating the given equation on both sides we get,
$\large\frac{dy}{dx} $$= ae^x + (-be^-x) + 2x$
Differenting again we get,
$\large\frac{d^2y}{dx^2} $$= ae^x +be^{-x} + 2$
Step 2:
Now substitute for $x, \large\frac{dy}{dx}$ and $\large\frac{d^2y}{dx^2}$
$x(ae^x +be^{-x} +2) + 2(ae^x - be^{-x} +2x) - x(ae^x + be^{-x} + x^2) + x^2 +2$
$xae^x +bxe^{-x} + 2x + 2ae^x - 2be^{-x} + 4x -xae^x -be^{-x} -x^2 +x^2 +2$ which is not equal to 0
Hence this is not a solution to the given differential equation.
answered Jul 30, 2013 by sreemathi.v
 

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