Question-94956

Question Number 94956 by i jagooll last updated on 22/May/20

Commented by niroj last updated on 22/May/20

both of you are right its note :

Commented by i jagooll last updated on 22/May/20

edit x \frac{d y}{d x} + 3 x (3 x + 1) y = e^{- 3 x}

Commented by niroj last updated on 22/May/20

x \frac{dy}{dx} + 3 x (3 x + 1) y = e^{- 3 x}

\frac{dy}{dx} + \frac{3 x (3 x + 1)}{x} y = \frac{e^{- 3 x}}{x}

\frac{dy}{dx} + 3 (3 x + 1) y = \frac{e^{- 3 x}}{x}

P = 3 (3 x + 1), Q = \frac{e^{- 3 x}}{x}

IF = e^{\int Pdx} = e^{\int 3 (3 x + 1) dx}

= e^{9 \int xdx + 3 \int dx}

= e^{9 \times \frac{x^{2}}{2} + 3 x}

= e^{\frac{9}{2} x^{2}} . e^{3 x}

y . e^{\frac{9}{2} x^{2}} . e^{3 x} = \int e^{\frac{9}{2} x^{2}} . e^{3 x} . \frac{e^{- 3 x}}{x} dx + C

y e^{\frac{9}{2} x^{2} + 3 x} = \int \frac{1}{x} e^{\frac{9}{2} x^{2}} dx + C

y = \frac{1}{e^{\frac{9}{2} x^{2} + 3 x}} [\int \frac{1}{x} e^{\frac{9}{2} x^{2}} dx + C] . .

now you can runway . . free . . ly

Commented by bobhans last updated on 22/May/20

hahaha��

Commented by bobhans last updated on 22/May/20

i think in 9^{th} line is wrong . it

should be \int \frac{e^{\frac{9}{2} x^{2}}}{x} d x

Commented by i jagooll last updated on 22/May/20

how \int \frac{e^{\frac{9}{2} x^{2}} d x}{x} \Rightarrow \int {x e}^{\frac{9}{2} x^{2}} d x s i r

Commented by peter frank last updated on 22/May/20

thank you

Answered by mathmax by abdo last updated on 22/May/20

{xy}^{^{'}} + 3 x (3 x + 1) y = e^{- 3 x}

(he) \to {xy}^{^{'}} = - 3 x (3 x + 1) y \Rightarrow \frac{y^{'}}{y} = - 3 (3 x + 1) \Rightarrow

\ln ∣ y ∣ = - 3 \int (3 x + 1) dx = - 3 (\frac{3}{2} x^{2} + x) = - \frac{9}{2} x^{2} - 3 x + c \Rightarrow

y = k e^{- \frac{9}{2} x^{2} - 3 x} mvc method \Rightarrow y^{^{'}} = (- 9 x - 3) k e^{- \frac{9}{2} x^{2} - 3 x}

+ k^{^{'}} e^{- \frac{9}{2} x^{2} - 3 x}

(e) \Rightarrow (- {9 x}^{2} - 3 x) e^{- \frac{9}{2} x^{2} - 3 x} + {xk}^{^{'}} e^{- \frac{9}{2} x^{2} - 3 x} + ({9 x}^{2} + 3 x) k e^{- \frac{9}{2} x^{2} - 3 x}

= e^{- 3 x} \Rightarrow {xk}^{^{'}} e^{- \frac{9}{2} x^{2} - 3 x} = e^{- 3 x} \Rightarrow k^{'} = \frac{1}{x} e^{\frac{9}{2} x^{2}} \Rightarrow

k (x) = \int_{.}^{x} \frac{1}{t} e^{\frac{9}{2} t^{2}} dt + c \Rightarrow y (x) = (\int_{.}^{x} \frac{1}{t} e^{\frac{9}{2} t^{2}} dt + c) e^{- \frac{9}{2} x^{2} - 3 x}