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

dy dx =aP e ^ ax +bQ e ^ bx ⟶ eq.1 ⇒ (a+b) dy dx =( a b)(aP e ^ ax +bQ e ^ bx ) ⇒ a ^ 2 P e ^ ax abQ e ^ bx abP e ^ ax b ^ 2 Q e ^ bx ⟶ ...

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Formation of a Differential Equation from a General Solution

If y = Peax...

Question

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

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\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

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