Question

The differential equation of the family of curves $y=e^{2x}(a\cos x+b\sin x)$, where a and b are arbitrary constants, is given by

• A. $y_2-4y_1+5y=0$
• B. $2y_2-y_1+5y=0$
• C. $y_2+4y_1-5y=0$
• D. $y_2-2y_1+5y=0$

Answer 1

Answer: (A)

$y_2-4y_1+5y=0$

Since, $y=e^{2x}(a\cos x+b\sin x)$
On differentiating w.r.t.x, we get
$\frac{dy}{dx}=e^{2x}(-a\sin x+b\cos x)+(a\cos x+b\sin x){2e}^{2x}$
$\frac{dy}{dx}=e^{2x}(-a\sin x+b\cos x)+2y$
Again differentiating, we get
$\frac{d^{2}y}{{dx}^2}=e^{2x}(-a\cos x-b\sin x)+(-a\sin x+b\cos x)e^{2x}.2+2\frac{dy}{dx}$
$=-e^{2x}(a\cos x+b\sin x)+{2e}^{2x}(-a\sin x+b\cos x)+2\frac{dy}{dx}$
$=-y+{2e}^{2x}(-a\sin x+b\cos x)+2\frac{dy}{dx}$
$\therefore y_2-4y_1+5y$
$=-y+2\frac{dy}{dx}+{2e}^{2x}(-a\sin x+b\cos x)-4\frac{dy}{dx}+5y$
$=4y-2\frac{dy}{dx}+{2e}^{2x}(-a\sin x+b\cos x)$
$=4y-{2e}^{2x}(-a\sin x+b\cos x)-4y+{2e}^{2x}(-a\sin x+b\cos x)$
$=0$
Hence, $y_2-4y_1+5y=0$