Find a particular solution Satisfying the given condition y = ae2x + be-x

Asked by Aaryan | 1 year ago |  61

##### Solution :-

From the question it is given that,

y = ae2x + be-x … [equation (i)]

Now, differentiate the equation (i) with respect x,

$$\dfrac{ dy}{dx}$$= 2 ae2x – be-x … [equation (ii)]

Then, the above equation is again differentiating with respect to x we get,

$$\dfrac{ d^2y}{dx^2}$$= 4 ae2x + be-x … [equation (iii)]

The given differential equation is $$\dfrac{ d^2y}{dx^2}$$$$\dfrac{ dy}{dx}$$ – 2y = 0

Substitute the equation (i), equation (ii) and equation (iii) in given differential equation,

$$\dfrac{ d^2y}{dx^2}$$ – $$\dfrac{ dy}{dx}$$ – 2y = 0

(4 ae2x + be-x) – (2 ae2x – be-x) – 2(ae2x + be-x) = 0

4 ae2x + be-x – 2ae2x + be-x – 2ae2x – 2be-x = 0

4 ae2x – 4ae2x + 2be-x – 2be-x = 0

0 = 0

Hence it is verified that, y = ae2x + be-x is a solution of the differential equation is $$\dfrac{ d^2y}{dx^2}$$ – $$\dfrac{ dy}{dx}$$ – 2y = 0.

Answered by Aaryan | 1 year ago

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