If $y=P e^{ax}+Q e^{bx}$, show that $\dfrac{d^{2}y}{dx^{2}}-(a+b)\dfrac{dy}{dx}+ab\,y=0$. (2014)

Differentiation — CBSE Class 12 · 2014 Board Question with Solution

y = P e^{a x} + Q e^{b x}

, show that

\frac{d ^{2} y}{d x ^{2}} - (a + b) \frac{d y}{d x} + ab y = 0

. (2014)

Answer: Both $e^{a x}$ and $e^{b x}$ coefficients vanish, so the relation holds (proved).

Given

y = P e^{a x} + Q e^{b x}

.
First derivative:

\frac{d y}{d x} = a P e^{a x} + b Q e^{b x} .

Second derivative:

\frac{d ^{2} y}{d x ^{2}} = a^{2} P e^{a x} + b^{2} Q e^{b x} .

Form the required combination:

\frac{d ^{2} y}{d x ^{2}} - (a + b) \frac{d y}{d x} + ab y .

Collect the

P e^{a x}

terms:

a^{2} - (a + b) a + ab = a^{2} - a^{2} - ab + ab = 0

.
Collect the

Q e^{b x}

terms:

b^{2} - (a + b) b + ab = b^{2} - ab - b^{2} + ab = 0

.
Both coefficients vanish, so the whole expression is

0

\frac{d ^{2} y}{d x ^{2}} - (a + b) \frac{d y}{d x} + ab y = 0

CBSE Class 12 Board · 2014 · Subjective