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. $y_2 - 4 y_1 + 5y =0$
• 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)

$y_2 - 4 y_1 + 5y =0$

Answer 2

Since, $y=e^{2 x}(a \cos x+b \sin x)$
On differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=e^{2 x}(-a \sin x +b \cos x)$
$+(a \cos x+ b \sin x) 2 e^{2 x}$
$\frac{d y}{d x}=e^{2 x}(-a \sin x +b \cos x)+2 y$
Again differentiating, we get
$\frac{d^{2} y}{d x^{2}}= e^{2 x}(-a \cos x-b \sin x)$
$+(-a \sin x +b \cos x) e^{2 x} \cdot 2+2 \frac{d y}{d x}$
$=-e^{2 x}(a \cos x +b \sin x)$
$+2 e^{2 x}(-a \sin x +b \cos x)+2 \frac{d y}{d x}$
$=-y +2 e^{2 x}(-a \sin x +b \cos x)+2 \frac{d y}{d x}$
$\therefore y_{2}-4 y_{1}+5 y$
$=-y+2 \frac{d y}{d x}+2 e^{2 x}(-a \sin x+ b \cos x)$
$-4 \frac{d y}{d x}+5 y$
$=4 y-2 \frac{d y}{d x}+2 e^{2 x}(-a \sin x +b \cos x)$
$=4 y-2 e^{2 x}(-a \sin x +b \cos x)-4 y$
$+2 e^{2 x}(-a \sin x +b \cos x)$
$=0$
Hence, $y_{2}-4 y_{1}+5 y=0$