Solve the equations $$x\dfrac{dy}{dx} = x + y$$

Asked by Sakshi | 1 year ago |  79

##### Solution :-

From the question it is given that,

$$x(\dfrac{dy}{dx}) = x + y$$

$$\dfrac{dy}{dx}=\dfrac{x+y}{x}$$

$$\dfrac{dy}{dx} = 1 + (\dfrac{y}{x})$$ … [equation (i)]

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

So, p = $$\dfrac{-1}{x}$$, Q = 1

$$IF = e^{∫pdx}$$

= $$e^{∫\dfrac{-1}{xdx}}$$

= e-log x

= $$e^{log(\dfrac{1}{x})}$$

$$\dfrac{1}{x}$$

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

$$Y(\dfrac{1}{x}) = ∫1 (\dfrac{1}{x}) dx + c$$

yex = log x + c

y = x log x + cx

Answered by Aaryan | 1 year ago

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