Answer to Question #284291 in Differential Equations for Aysu

Question #284291

 Solve the first order linear inhomogeneous differential equation using the Bernoulli method

y,-(2y/x)=1+(1/x)


1
Expert's answer
2022-02-09T16:41:17-0500

Bernoulli equation is of the form "\\displaystyle\n\\frac{dy}{dx}+p(x)y=q(x)y^n". The given DE has n=0 which reduces the standard Bernoulli equation to "\\displaystyle\n\\frac{dy}{dx}+p(x)y=q(x)" which can be solved as a first order linear differential equation.


Now, the given DE is;

"\\displaystyle\ny\\prime-\\left(\\frac{2}{x}\\right)y=1+\\frac{1}{x}"

putting "y=uv" into the given DE yields;

"\\displaystyle\nuv\\prime+vu\\prime-\\left(\\frac{2}{x}\\right)uv=1+\\frac{1}{x}\\\\\n\\Rightarrow uv\\prime+v\\left[u\\prime-\\frac{2}{x}u\\right]=1+\\frac{1}{x}\\cdots\\cdots(1)\\\\"

Putting the coefficient of "v" equal zero yields;

"\\displaystyle\nu\\prime-\\frac{2}{x}u=0\\Rightarrow\\frac{du}{dx}=\\frac{2u}{x}"

By method of separation of variables, we have;

"\\displaystyle""\\displaystyle\n\\frac{1}{u}\\frac{du}{dx}=\\frac{2}{x}\\\\\n\\Rightarrow\\int\\frac{1}{u}\\frac{du}{dx}\\ dx=\\int\\frac{2}{x}\\ dx\\Rightarrow \\int\\frac{1}{u}\\ du=\\int\\frac{2}{x}\\ dx\\Rightarrow \\ln u=\\ln x^2+\\ln c\\\\\n\\Rightarrow u=cx^2, \\text{where c is an arbitrary constant.}"


Next substituting "u=cx^2" into (1) yields;

"\\displaystyle\ncx^2v\\prime=1+\\frac{1}{x}"

By method of separation of variables, we have;

"\\displaystyle\nc\\frac{dv}{dx}=\\frac{1}{x^2}+\\frac{1}{x^3}"


integrating both sides wrt "x" yields;

"\\displaystyle\nc\\int\\frac{dv}{dx}\\ dx=\\int\\left(\\frac{1}{x^2}+\\frac{1}{x^3}\\right)\\ dx\\Rightarrow c\\int dv=\\int\\left(\\frac{1}{x^2}+\\frac{1}{x^3}\\right)\\ dx"

"\\displaystyle\ncv=-\\frac{1}{x}-\\frac{1}{2x^2}+k\\Rightarrow v=\\frac{1}{c}\\left(-\\frac{1}{x}-\\frac{1}{2x^2}+k\\right)" , where k is an arbitrary constant.

Thus,

"\\displaystyle\ny=uv=cx^2\\times \\frac{1}{c}\\left(-\\frac{1}{x}-\\frac{1}{2x^2}+k\\right)=-x-\\frac{1}{2}+kx^2=kx^2-x-\\frac{1}{2}"

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