The differential equation of y=e^{x}≤ft(a+{bx}+x²) is

Correct Answer is : d2ydx2−2dydx−2ex+y=0{d² y}{{dx}²}-2 d y/d x-2 e^{x}+y=0dx2d2y−2dxdy−2ex+y=0

{d² y}{{dx}²}-2 d y/d x-2 e^{x}+y=0

{d² y}{{dx}}+2 d y/d x-e^{x}+2 y=0

Correct Answer is : d2ydx2−2dydx−2ex+y=0{d² y}{{dx}²}-2 d y/d x-2 e^{x}+y=0dx2d2y−2dxdy−2ex+y=0

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MHTCET 15 May 2024 Shift 1 Paper

Question : 148 of 150

Marks: +1, -0

y=e^{x}\left(a+\mathrm{bx}+x^{2}\right)

Solution:

Concept:

The given function contains two arbitrary constants

a

and

b

, so differentiate twice and eliminate them to form the differential equation.

Explanation:

Let

y = e^{x}(a + bx + x^{2})

Differentiate with respect to

x

\frac{dy}{dx} = e^{x}(a + bx + x^{2}) + e^{x}(b + 2x)

Since

e^{x}(a + bx + x^{2}) = y

, we get:

\frac{dy}{dx} = y + e^{x}(b + 2x)

... (i)

Differentiate (i) again:

\frac{d^{2}y}{dx^{2}} = \frac{dy}{dx} + e^{x}(b + 2x) + 2e^{x}

Using (i),

e^{x}(b + 2x) = \frac{dy}{dx} - y

, so:

\frac{d^{2}y}{dx^{2}} = \frac{dy}{dx} + \left(\frac{dy}{dx} - y\right) + 2e^{x}

Simplify and rearrange:

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

Therefore, the required differential equation is:

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

Answer:

Option C:

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

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