Question

(i) Form the DE corresponding to the Function y = ae^x + be^2x

(ii) State the order and degree of the above DE.

(iii) Solve x\(\frac{dy}{dx}\) = x + y

Answer 1

(i) y = ae^x + be^2x ..........(1)

\(\frac{dy}{dx} \) =  ae^x + 2be^2x  ........(2)

\(\frac{d^2y}{dx^2} \) = ae^x + 4be^2x ..........(3)

(3) - 3(2) + 2 (1)

(ii) order : 2 and Degree : 1

(iii) x\(\frac{dy}{dx}\) = x + y ⇒ \(\frac{dy}{dx}\) = \(\frac{x+y}{x}\)

This is Homogeneous DE.

Put y = vx and \(\frac{dy}{dx} = v + \frac{dv}{dx}\) 

This is a Homogeneous DE.

Put y = vx and \(\frac{dy}{dx} = \frac{x + vx}{x}\)

⇒ x\(\frac{dv}{dx} = 1 + v-v ⇒ dv = \frac{dx}{x}\)

Integrating on both sides,