Solve the equations $$4\dfrac{dy}{dx} + 8y = 5e^{-3x}$$

Asked by Sakshi | 1 year ago |  109

##### Solution :-

From the question it is given that,

4($$\dfrac{dy}{dx}$$) + 8y = 5e-3x

Dividing both side by 4 we get,

$$\dfrac{dy}{dx}$$ + 2y = $$\dfrac{ 5e^{-3x}}{4}$$ … [equation (i)]

The given linear differential equation is comparing with, ($$\dfrac{d}{dx}$$) + py = Q

So, p = 2, Q = $$\dfrac{ 5e^{-3x}}{4}$$

IF = e∫pdx

= e∫2dx

= e2x

Then, multiplying both side of equation (i) by IF,

e2x($$\dfrac{d}{dx}$$) + e2x 2y = e2x × $$\dfrac{ 5e^{-3x}}{4}$$

e2x($$\dfrac{d}{dx}$$) + e2x 2y = $$\dfrac{ 5e^{-x}}{4}$$

Now, integrating the above equation with respect to x,

ye2x = $$\int \dfrac{ 5e^{-3x}}{4}$$ dx+ c

ye2x = ($$\dfrac{ -5}{4}$$) e-x + c

y = ($$\dfrac{ -5}{4}$$) e-3x + ce-2x

Answered by Aaryan | 1 year ago

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