How can i solve this differencial equation?: y ′ + x 2 y = x 2

Question

How can i solve this differencial equation?:       y    ′    +      x          2        y  =      x          2

Giancarlo Shah · Accepted Answer

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

\frac{d y}{d x} + P (x) y = Q (x)

We have:

y^{'} = x^{2} y = x^{2}

...[1]
This is a First Order Ordinary Differential Equation in Standard Form. So we compute and integrating factor, I, using;

I = e^{\int P (x) d x}

= \exp (\int x^{2} d x)

= \exp (\frac{1}{3} x^{3})

=e^{\fra

= e^{\frac{1}{3} x^{3}}

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

y^{'} e^{\frac{1}{3} x^{3}} + x^{2} e^{\frac{1}{3} x^{3}} y = x^{2} e^{\frac{1}{3} x^{3}}

∴ \frac{d}{d x} (y e^{\frac{1}{3} x^{3}}) d x + C

Leading to the explicit General Solution:

y = e^{- \frac{1}{3} x^{3}} {e^{\frac{1}{3} x^{3}} + C}

= 1 + C e^{- \frac{1}{3} x^{3}}

vilitatelp014 · Accepted Answer

y    ′    +      x          2        y  =      x          2      This is a separable ODE            d      y              d      x        =      x          2        −      x          2        y  =      x          2        (  1  −  y  )  ⇒  ∫      1          1      −      y        d  y  =  ∫      x          2        d  x  ⇒  −  ln  ⁡      |    1  −  y      |    =      1    3        x          3        +  c  1  −  y  =  B      e                  1        3                    x                  3                     where   B  =      e          −      c        y  =  1  +  A      e                  1        3                    x                  3                     where A=-B

How can i solve this differencial equation?: y ′ </msup> + x <mrow

Answered question

Answer & Explanation

New Questions in Calculus 2