Y-aex-be-x-x2-xd2y-dx2-2dy-dx-xy-x2-2-0

We have, y=ae^x+be^ x+x^2 ....... ( #10;1 ) #10; #10;On differentiating this #10;with respect to x and we get, #10; #10; dydx=ddx( #10 ...

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Byju's Answer

Standard XII

Mathematics

Formation of a Differential Equation from a General Solution

y=aex+be-x+x2...

Question

$y = a e^{x} + b e^{- x} + x^{2} : x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 = 0$

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Solution

We have,
$y = a e^{x} + b e^{- x} + x^{2} . . . . . . . (1)$

On differentiating this with respect to x and we get,

$\frac{d y}{d x} = \frac{d}{d x} (a e^{x} + b e^{- x} + x^{2})$

$\frac{d y}{d x} = a e^{x} - b e^{- x} + 2 x . . . . . . . (2)$

Again, differentiating and we get,
$\frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (a e^{x} - b e^{- x} + 2 x)$

$= a e^{x} + b e^{- x} + 2 . . . . . . . (3)$

Now given equation.
$x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 = 0$

By equation (1), (2) and (3) value put in given equation.

$\Rightarrow x (a e^{x} + b e^{- x} + 2) + 2 (a e^{x} - b e^{- x} + 2 x) - x (a e^{x} + b e^{- x} + x^{2}) + x^{2} - 2 = 0$

$\Rightarrow a x e^{x} + b x e^{- x} + 2 x + 2 a e^{x} - 2 b e^{- x} + 4 x - a x e^{x} - b x e^{- x} - x^{3} + x^{2} - 2 = 0$

$\Rightarrow 2 a e^{x} - 2 b e^{- x} + 6 x - x^{3} + x^{2} - 2 = 0$
Hence, this is the general solution.

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Similar questions

For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

(i)

(ii)

(iii)

(iv)

Q. For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

y = a e^{x} + b e^{- x} + x^{2}

x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 = 0

For the question given below verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.

$y = a e^{x} + b e^{- x} + x^{2}; x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 = 0$

Q. Assertion :The order of the differential equation, of which

x y = c e^{x} + b e^{- x} + x^{2}

is a solution, is

2

Reason: The differential equation is

x \frac{d^{2} y}{{d x}^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 = 0

Q. If

y = x^{3} log (\frac{1}{x})

, then prove that

x \frac{d^{2} y}{d x^{2}} - \frac{2 d y}{d x} + 3 x^{2} = 0

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y=aex+be−x+x2:xd2ydx2+2dydx−xy+x2−2=0

$y = a e^{x} + b e^{- x} + x^{2} : x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 = 0$