Question

It is easy to check that for any value of c, the function y=x2+cx2 is solution of equation xy′+2y=4x2, (x>0). Find the value of c for which the solution satisfies the initial condition y(10)=2.

Answer 1

Answer:

Therefore $y=x^2+\frac{c}{x^2}$ is a solution of $xy'+2y=4x^2$.

$\therefore y=x^2-\frac{9800}{x^2}$

Step-by-step explanation:

The given differential equation is

$xy'+2y=4x^2$

$\Rightarrow y'+\frac{2y}{x}=4x$

Here $P=\frac {2}{x}$  and Q= 4x

The integrating factor is $=e^{\int p dx$

                                       $=e^{\int \frac{2}{x} dx$

                                       $=e^{2\ ln x}$

                                       $= e^{lnx^2}$

                                       $=x^2$

Multiplying the integrating factor both sides of the DE

$x^2y'+x^2\frac{2y}{x}=4x.x^2$

$\Rightarrow x^2 \frac{dy}{dx}+2yx=4x^3$

$\Rightarrow x^2dy+2yxdx=4x^3dx$

Integrating both sides

$\int x^2dy+\int 2yxdx=\int 4x^3dx$

$\Rightarrow x^2y= \frac{4x^4}{4}+c$          [ c is an arbitrary constant]

$\Rightarrow x^2y =x^4+c$    

$\Rightarrow y=x^2+\frac{c}{x^2}$             [ dividing by x²]

Therefore $y=x^2+\frac{c}{x^2}$ is a solution of $xy'+2y=4x^2$.

The initial condition is y(10)=2

Putting x=10 and y=2

$\therefore 2=10^2+\frac{c}{10^2}$

$\Rightarrow 200= 10000+c$   [ multiply by 100]

$\Rightarrow c=200-10000$

$\Rightarrow c=-9800$

Putting the value of c in the solution we get,

$\therefore y=x^2-\frac{9800}{x^2}$