If y - Peax - Qebx- show that d2ydx2 - a - b dydx - aby - 0-

y=Pe^ax+Qe^bx — (1) dydx=aPe^ax+bQe^bx — (2) d^2ydx^2=a^2Pe^ax+b^2Qe^bx — (3)Multiplying (1) by ab, we get aby=abPe^ax+abQe^bx —(4)Multiply ...

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Higher Order Derivatives

If y = Peax...

Question

If $y = P e^{a x} + Q e^{b x}$ , show that $\frac{d^{2} y}{d x^{2}} - (a + b) \frac{d y}{d x} + a b y = 0.$

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If $y = P e^{a x} + Q e^{b x}$ , then prove that $\frac{d^{2} y}{d x^{2}} - (a + b) \frac{d y}{d x} + a b y = 0$

y = cos (sin x)

\frac{d^{2} y}{d x^{2}} + tan x \frac{d y}{d x} + y {cos}^{2} x = 0

y = x^{2} e^{x}

\frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} - 2 (x + 1) e^{x} = 0

y = {sin}^{- 1} x

(1 - x^{2}) . \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = 0

y = A e^{m x} + B e^{n x}

\frac{d^{2} y}{d x^{2}} - (m + n) \frac{d y}{d x} + m n y = 0

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