How do you solve $xy'+2y=4x^2$ given y(1)=0?

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Answer 1 · 2017-05-09T00:48:55

$y = x^2 -1/x^2$

We have:

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

Which we can write as:

$y'+2/xy=4x \ \ \ \ ...... [1]$

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

$dy/dx + P(x)y=Q(x)$

Then the integrating factor is given by;

$IF = e^(int P(x) dx)$
$" " = exp(int \ 2/x \ dx)$
$" " = exp( 2lnx )$
$" " = e^( lnx^2 )$
$" " = x^2$

And if we multiply the DE [1] by this Integrating Factor, $IF$ , we will have a perfect product differential;

$y'+2/xy=4x$

$:. x^2y'+2xy=4x^3$

$:. d/dx (x^2y) = 4x^3$

Which we can directly integrate to get:

$int \ d/dx (x^2y) \ dx = int \ 4x^3 \ dx$
$:. x^2y = x^4 + C$

Applying the initial condition, $y(1)=0$ , we get:

$1*0 = 1 + C => C = -1$

Thus the solution is:

$\ x^2y = x^4 -1$
$:. y = x^2 -1/x^2$

How do you solve xy'+2y=4x^2xy'+2y=4x^2 given y(1)=0?