Answer :
[tex]y=\dfrac{4\ln x+C}x\implies y'=\dfrac{x(4\ln x+C)'-(4\ln x+C)(x)'}{x^2}=\dfrac{4-4\ln x-C}{x^2}[/tex]
Substitute this into the differential equation:
[tex]x^2y'+xy=4-4\ln x-C+4\ln x+C=4[/tex]
so every solution of the given form [tex]y[/tex] is a solution.
Thus, the every member of the family of functions is a solution of the provided differential equation as left hand side of equation is equal to the right hand side.
What is differential equation?
The differential equation is the function in which minimum one darivative of unknown function exists.
The differential equation given in the problem is,
[tex]x^2y' + xy = 4[/tex] .......1
This solution of this differential equation is,
[tex]y =\dfrac{4\ln(x) + C}{x}[/tex]
Differenciate the above equation, with respect to the x,
[tex]y' =\dfrac{x(4\ln(x) + C)-(4\ln(x) + C)x'}{x^2}\\y' =\dfrac{4-4\ln(x) - C)}{x^2}[/tex]
Put this value in the equation number 1 as,
[tex]x^2(\dfrac{4-4\ln(x) - C)}{x^2} ) + x\left(\dfrac{4\ln (x)+C}{x}\right) = 4\\4-4\ln(x) - C+ 4\ln (x)+C = 4\\4=4[/tex]
Thus, the every member of the family of functions is a solution of the provided differential equation as left hand side of equation is equal to the right hand side.
Learn more about differential equation here;
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