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

satisfying y(1) = 0

Question

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

satisfying y(1) = 0

myininaya · Answer

it looks like we can write this in the form:
$$y'+p(x)y=q(x)$$

$$xy'=x^2e^{-x}+y =>xy'-y=x^2e^{-x}=> y'-\frac{1}{x}y=xe^{-x}$$

myininaya · Answer

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
$$(\frac{1}{x}y)'=e^{-x}$$

myininaya · Answer

now we need to integrate both sides

myininaya · Answer

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

myininaya · Answer

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

myininaya · Answer

$$C=e^{-1}=\frac{1}{e}$$

myininaya · Answer

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

anonymous · Answer

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:
$$y = C _{1}x-xe ^{-x}$$

anonymous · Answer

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

myininaya · Answer

formulas are too easy to use lol

anonymous · Answer

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

anonymous · Answer

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

myininaya · Answer

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