Tentukan solusi persamaaan diferensial (6x + y) dx+ (4x + y) dy = 0

Tentukan solusi persamaaan diferensial (6x + y) dx+ (4x + y) dy = 0

hasil integralnya

$3{x}^{2} + y + 4x + frac{1}{2} {y}^{2} + c = 0$

Materi: Persamaan Diferensial Homogen

Jawaban:

$4x^2+y^2+4xy-3Cx-Cy=0$

Penjelasan dengan langkah-langkah:

(6x+y) dx + (4x+y) dy = 0 (bagi kedua ruas dengan x)

$left(6+frac{y}{x}right)dx+left(4+frac{y}{x}right)dy=0\left(6+frac{y}{x}right)+left(4+frac{y}{x}right)frac{dy}{dx}=0$

Misalkan:

u = y/x —-> y = xu

dy/dx = u + x du/dx

Substitusikan ini ke PD, sehingga:

$tiny{begin{aligned}\left(6+frac{y}{x}right)+left(4+frac{y}{x}right)frac{dy}{dx}&=0\(6+u)+(4+u)left(u+xfrac{du}{dx}right)&=0\(6+u)+4u+4xfrac{du}{dx}+u^2+xufrac{du}{dx}&=0\6+5u+u^2&=-x(4+u)frac{du}{dx}\frac{u+4}{u^2+5u+6}du&=-frac{1}{x}dx\int{frac{u+4}{u^2+5u+6},du}&=-int{frac{1}{x},dx}\int{frac{u+4}{(u+3)(u+2)},du}&=-ln{x}+Cend{aligned}}$

Untuk mencari integral di ruas kiri, gunakan pecahan parsial, sehingga:

$tiny{begin{aligned}\int{frac{u+4}{(u+3)(u+2)},du}&=-ln{x}+C\int{left(frac{2}{u+2}-frac{1}{u+3}right), du}&=ln{x^{-1}}+C\2ln{(u+2)}-ln{(u+3)}=ln{x^{-1}}+ln{C}\ln{left(frac{{(u+2)}^2}{u+3}right)}&=ln{(Cx^{-1})}\frac{u^2+4u+4}{u+3}&=Cx^{-1},(times,(u+3))\u^2+4u+4=Cx^{-1}(u+3)end{aligned}}$

Karena u = y/x, maka:

$tiny{begin{aligned}\left(frac{y^2}{x^2}right)+4left(frac{y}{x}right)+4&=Cx^{-1}left(frac{y}{x}+3right),(times,x^2)\y^2+4xy+4x^2&=Cy+3Cx\4x^2+y^2+4xy-3Cx-Cy&=0end{aligned}}$

Semoga membantu.