Question

Find solutions to:
x(dy/dx)=(x^2)*(e^-x) + y

satisfying y(1) = 0

Answer 1

it looks like we can write this in the form:
\[y'+p(x)y=q(x)\]

\[xy'=x^2e^{-x}+y =>xy'-y=x^2e^{-x}=> y'-\frac{1}{x}y=xe^{-x}\]

Answer 2

now this is pretty easy to solve
first thing we need to do is multiply both sides by
\[v(x)=e^{\int\limits_{}^{}-\frac{1}{x} dx}=e^{-lnx}=\frac{1}{x}\]
\[\frac{1}{x}y'-\frac{1}{x} \cdot \frac{1}{x} y=e^{-x}\]
we did so we could write
\[(\frac{1}{x}y)'=e^{-x}\]

Answer 3

now we need to integrate both sides

Answer 4

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

Answer 5

\[\frac{1}{1} (0)=-e^{-1}+C\] after pluggin' x=1 and y=0 we can solve for C

Answer 6

\[C=e^{-1}=\frac{1}{e}\]

Answer 7

so we have
\[\frac{1}{x}y=-e^{-x}+\frac{1}{e}\]

Answer 8

That is a good solution using the integration constant \[v(x) = e ^{\int\limits_{}^{}-p(x)dx}\]

Additionally you can use an equation that is also derived from the integration constant for the general form to give you the solution.\[y \prime+p(x)y=q(x)\]

\[y = \ \frac{1}{v(x)} \ \int\limits_{}^{}q(x)v(x)dx+C\]

By the way the final step for your solution will be:
\[y = C _{1}x-xe ^{-x}\]

Answer 9

Okay yeah and the value of constant \[C _{1}=\ \frac{1}{e} \]

Answer 10

formulas are too easy to use lol

Answer 11

Yes you're right..I particularly don't go that route but I like it because my prof. did a great job deriving it from general formula and some people like using it too

Answer 12

I don't mind using formulas that I can derive completely it makes more sense that way.

Answer 13

i like doing it long way because it makes more sense to me why that is the answer