Expert Answer to "Solve x3y+x2y−2xy+2y=x3ln⁡x"

Name: Solve x3y+x2y−2xy+2y=x3ln⁡x
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solve x3y+x2y−2xy+2y=x3ln⁡x

Emeli Hagan

Answered question

2020-12-09

Solve

x^{3} y + x^{2} y - 2 x y + 2 y = x^{3} \ln x

Answer & Explanation

Mayme

Skilled2020-12-10Added 103 answers

Heres

Jeffrey Jordon

Expert2021-09-30Added 2605 answers

Given: $x^{3} y + x^{2} y - 2 x y + 2 y = x^{3} \ln x$

Let $\ln x = z$ then

$\frac{d y}{d x} = \frac{d y}{d z} \frac{d z}{d x} = \frac{1}{x} \frac{d y}{d x}$

Similarly:

$x^{2} \frac{d^{2} y}{d x^{2}} = \frac{d^{2} y}{d z^{2}} - \frac{d y}{d z}$ and $x^{3} \frac{d^{3} y}{d x^{3}} = \frac{d^{3} y}{d z^{3}} - 3 \frac{d^{2} y}{d z^{2}} + 2 \frac{d y}{d x}$

Using above three expressions, the original differential equation reduces to:

$(\frac{d^{3} y}{d z^{3}} - 3 \frac{d^{2} y}{d z^{2}} + 2 \frac{d y}{d z}) + (\frac{d^{2} y}{d z} - \frac{d y}{d z}) - 2 \frac{d y}{d z} + 2 y = z e^{3 z}$

The auxilary equation of the differential equation

$\frac{d^{3} y}{d z^{3}} - 2 \frac{d^{2} y}{d z^{2}} - \frac{d y}{d z} + y = 0$ becomes

$m^{3} - 2 m^{2} - m + 2 = 0$

$\Rightarrow m = 2, 1, - 1$

Therefore the complimentary function becomes:

$y_{c} (x) = a e^{2 z} + b e^{z} + c e^{- z}$

The potential integral is given by:

$y_{p} (x) = \frac{1}{D^{3} - 2 D^{2} - D + 2} z e^{3 z}$

$= e^{3 z} \frac{1}{(D + 3)^{3} - 2 (D + 3)^{2} - (D + 3) + 2} z$

$= e^{3 z} \frac{1}{D^{3} + 7 D^{2} + 14 D + 4} z$

$= e^{3 z} \frac{1}{4} (1 + \frac{D^{3} + 7 D^{2} + 14 D}{4})^{- 1} z$

$= \frac{e^{3 z}}{4} (1 - \frac{D^{3} + 7 D^{2} + 14 D}{4} + . . .) z$

$= \frac{e^{3 z}}{4} (z - \frac{14}{4})$

$= \frac{x^{3}}{4} (\ln x - \frac{14}{4})$

Therefore the solution is

$y (x) = y_{c} (x) + y_{p} (x) = a x^{2} + b x + \frac{c}{x} + \frac{x^{3}}{4} (\ln x - \frac{14}{4})$

where a,b,c are arbitrary constants

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