Find a particular solution Satisfying the given condition y = bex + ce2x

Asked by Aaryan | 1 year ago |  56

##### Solution :-

From the question it is given that,

y = bex + ce2x … [equation (i)]

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

$$\dfrac{ dy}{dx}$$ = bex + 2ce2x … [equation (ii)]

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

$$\dfrac{ d^2y}{dx^2}$$= bex + 4ce2x … [equation (iii)]

The given differential equation is $$\dfrac{ d^2y}{dx^2}$$ – 3 ($$\dfrac{ dy}{dx}$$) + 2y = 0

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

$$\dfrac{ d^2y}{dx^2}$$ – 3 ($$\dfrac{ dy}{dx}$$) + 2y = 0

(bex + 4ce2x) – 3 (bex + 2ce2x) + 2(bex + ce2x) = 0

bex + 4ce2x – 3bex – 6ce2x + 2bex + 2ce2x = 0

3bex – 3bex + 6ce2x – 6ce2x = 0

0 = 0

Hence it is proved that, d2y/dx2 – 3 (dy/dx) + 2y = 0

Answered by Aaryan | 1 year ago

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