Question

Solve:
i) $y=e^{2x}\left(a+bx\right)$ $y=e^x\left(a\cos x+b\sin x\right)$
ii) Form the differential equation of the family of circle touching the y-axis at origin.

Answer 1

Part (1) let, $y=e^x(a+bx)$ …(1)

Differentiate with respect to x

$\frac{dy}{dx}=e^{2x}(0+b)+(a+bx)e^{2x}.2$

From equation (1)

$\frac{dy}{dx}=e^{2x}\left(b\right)+y.2$ …..(2)

Differentiate again equation (2) with respect to x

$\frac{d^{2}y}{d{x}^2}=b.e^{2x}.2+.2\frac{dy}{dx}$

From equation (2)

$\frac{d^{2}y}{d{x}^2}=(\frac{dy}{dx}-2y).2+.2\frac{dy}{dx}$

$\frac{d^{2}y}{d{x}^2}=4\frac{dy}{dx}-4y$

Part (2)

Let, $y=e^x(a\cos x+b\sin x)$ ….(1)

Differentiate with respect to x

$\frac{dy}{dx}=e^x(-a\sin x+b\cos x)+(a\cos x+b\sin x).e^x$

From equation (1)

$\frac{dy}{dx}=e^x(-a\sin x+b\cos x)+y$ …..(2)

Differentiate again equation (2) with respect to x

$\frac{d^{2}y}{d{x}^2}=e^x(-a\sin x-b\sin x)+\frac{dy}{dx}$

$\frac{d^{2}y}{d{x}^2}=-e^x(a\sin x+b\sin x)+\frac{dy}{dx}$

From equation (1)

$\frac{d^{2}y}{d{x}^2}=-y+\frac{dy}{dx}$

$\frac{d^{2}y}{d{x}^2}-\frac{dy}{dx}+y=0$