Question

Find solutions to the equation:
xy′+2y=x3
such that y = 1 when x = 1

Answer 1

is that differential equation?

Answer 2

Yes

Answer 3

\[xy'+2y=x^3\] divide by x
\[y'+2y/x=x^2\]

Answer 4

\[y=u*v\]
\[y=u'*v+u*v'\]
Put into equation
\[u'*v+u*v'+2uv/x=x^2\]
\[u'*v+u*(v'+2v/x)=x^2\]

Answer 5

Let \[v'+2v/x=0\] (1) then equation \[u'v=x^2\] (2)

Answer 6

\[xy'+2y=x^3\]

multiply by x

\[x^2y'+2yx=x^4\]

\[\frac{d}{dx}(x^2y)=x^4\]

integrate both sides ... solve for y

Answer 7

\[x^2y=\frac{1}{5}x^5+c\]

\[y=\frac{1}{5}x^3+\frac{c}{x^2}\]

now solve for the initial condition

Answer 8

(1)\[dv/dx=-2v/x\]
\[\int\limits( dv/v)=-2\int\limits dx/x\]

Answer 9

\[\ln(v)=-2\ln(x)\]\[v=x^{-2}\]

Answer 10

Put \[v=x^{-2}\] into (2) \[u'v=x^2\]\[u'x^{-2}=x^2\]

Answer 11

\[u'=x^4\]\[u=x^5/5+C\]

Answer 12

\[y=u*v=(x^5/5+C)*x^{-2}=x^3/5+C/x^2\]

Answer 13

y = 1 when x = 1
Put into the solution y=1 and x=1
\[1=1/5+C\], then C=1-1/5=4/5
So solution
\[y=x^3/5+4/(5x^2)\]

Answer 14

Sorry, my laptop died when I was working through that, I was following until I got to:
Let
v′+2v/x=0

Answer 15

@ Tatiana_Stebko is correct.\[y[x]= \frac{4}{5 x^2}+\frac{x^3}{5}\]\[x y'[x]+2y[x]\text{= }x^3\]\[x \left(-\frac{8}{5 x^3}+\frac{3 x^2}{5}\right)+2 \left(\frac{4}{5 x^2}+\frac{x^3}{5}\right)=x^3 \]\[\frac{-8+3 x^5}{5 x^2}+\frac{2 \left(4+x^5\right)}{5 x^2}= x^3\]\[x^3 = x^3\]

Answer 16

Why can we let
v′+2v/x=0

Answer 17

The easiest way to do this problem is with an integrating factor.

Answer 18

http://en.wikipedia.org/wiki/Integrating_factor#Use_in_solving_first_order_ordinary_differential_equations