contestada

The functions y=x2+cx2 are all solutions of equation: xy′+2y=4x2, (x>0). Find the constant c which produces a solution which also satisfies the initial condition y(4)=4.

Respuesta :

Answer:

The required value of c is -192.

Step-by-step explanation:

We are given the differential equation:

[tex]xy'+2y=4x^2 \\(x>0)[/tex]

The solution to the given differential equation is:

[tex]y=x^2+\dfrac{c}{x^2}[/tex]

Initial condition:

[tex]y(4) = 4[/tex]

Putting the values, in the solution, we get,

[tex]4=(4)^2+\dfrac{c}{(4)^2}\\\\4 = 16+\dfrac{c}{(4)^2}\\\\\Rightarrow \dfrac{c}{(4)^2}=-12\\\Rightarrow c =-12(16)\\\Rightarrow c =-192[/tex]

Thus, the required value of c is -192.

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