Answer to Question #242798 in Mechanical Engineering for Elimar

Question #242798

find the general solution of the given DE (x - 2y)dx - (x - 2y + 1)dy = 0


1
Expert's answer
2021-09-28T00:57:51-0400

(x-2y)dx-(x-2y+1)dy=0

(x-2y)dx=(x-2y+1)dy

"\\frac{dy}{dx}=\\frac{x-2y}{x-2y+1}" This is equation (i)

Let x-2y=u "\\implies1-2\\frac{dy}{dx}=\\frac{du}{dx}"

"1-\\frac{du}{dx}=2\\frac{dy}{dx}"

"\\frac{dy}{dx}=\\frac{1}{2}-\\frac{1}{2}\\frac{du}{dx}" This is equation (ii)

Substituting equation (ii) into equation(i) we get,

"\\frac{1}{2}-\\frac{1}{2}\\frac{du}{dx}=\\frac{u}{u+1}"

"1-\\frac{du}{dx}=\\frac{2u}{2u+2}"

"\\frac{du}{dx}=1-\\frac{2u}{2u+2}"

"\\frac{du}{dx}=\\frac{2u+2-2u}{2u+2}=\\frac{2}{2u+2}"

"(2u+2)du=2dx"

Integrate both sides"\\int (2u+2)du=2\\int 1dx"

"u^{2}+2u=2x+C"

"(x-2y)^{2}+2(x-2y)=2x+C"

"x^{2}-4xy+4y^{2}+2x-4y=2x+C"

"4y^{2}-(4x+4)y+(x^{2}+C)=0"

"y=\\frac{(4x+4)\\pm \\sqrt{(-4x-4)^{2}-16(x^{2}+C}}{8}"

"y=\\frac{4(x+1)\\pm \\sqrt{16x^{2}+32x+16-16x^{2}-16C}}{8}"

"y=\\frac{4(x+1)\\pm \\sqrt{32x+16-16C}}{8}"

"y=\\frac{4(x+1)\\pm 4\\sqrt{2x+1-C}}{8}"

"y=\\frac{(x+1)\\pm \\sqrt{2x+1-C}}{2}"

The general solution of the DE is;

"y=\\frac{(x+1)\\pm \\sqrt{2x+1-C}}{2}"


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