find-a-maclaurine-series-solution-to-the-differential-equation-up-to-the-term-in-x-4-dy-dx-x-xy-if-y-1-when-x-0-

Question Number 88235 by Rio Michael last updated on 09/Apr/20

find a maclaurine series solution to the differential equation

up to the term in x^{4} .

\frac{d y}{d x} - x = x y if y = 1 when x = 0 .

Commented by niroj last updated on 11/Apr/20

\frac{dy}{dx} - x = xy if y = 1 when x = 0 .

by the help of linear diff . . {equ}^{n} will be . .

\frac{dy}{dx} - x - xy = 0

\frac{dy}{dx} - xy = x

P = - x & Q = x

IF = e^{\int Pdx} = e^{- \frac{x^{2}}{2}}

y \times IF = \int IF \times Qdx + C

y . e^{- \frac{x^{2}}{2}} = \int e^{- \frac{x^{2}}{2}} . xdx + C

put x^{2} = t

2 xdx = dt

xdx = \frac{dt}{2}

y . e^{- \frac{x^{2}}{2}} = \int e^{- \frac{t}{2}} . \frac{dt}{2} + C

y . e^{- \frac{x^{2}}{2}} = \frac{1}{2} \int e^{- \frac{t}{2}} dt + C

{ye}^{- \frac{x^{2}}{2}} = \frac{1}{2} . \frac{e^{- \frac{t}{2}}}{- \frac{1}{2}} + C

{ye}^{- \frac{x^{2}}{2}} = - . e^{- \frac{x^{2}}{2}} + C

{ye}^{- \frac{x^{2}}{2}} + e^{- \frac{x^{2}}{2}} = C

if y = 1 and x = 0

1 . e^{- \frac{0}{2}} + e^{\frac{- 0}{2}} = C

C = 1.1 + 1 = 2

∴ {ye}^{- \frac{x^{2}}{2}} + e^{- \frac{x^{2}}{2}} = 2

\frac{y + 1}{e^{\frac{x^{2}}{2}}} = 2

y = {2 e}^{\frac{x^{2}}{2}} - 1 / / .

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