Solve the equations $$\dfrac{dy}{dx} + y = e^{-2x}$$

Asked by Sakshi | 1 year ago |  70

Solution :-

From the question it is given that,

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

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

So, p = 1, Q = e-2x

IF = e∫pdx

= e∫1dx

= ex

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

ex($$\dfrac{dy}{dx}$$) + ex y = ex × e-2x

ex($$\dfrac{dy}{dx}$$) + ex y = e-x … [because am × an = am + n]

Now, integrating the above equation with respect to x,

yex = ∫e-x dx + c

yex = $$\dfrac{e^{-x}}{-1} + c$$

y = -e-2x + ce-x

Answered by Aaryan | 1 year ago

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