can anyone solve xdy/dx+2y=e^-x/x?

Answer 1

dy/dx + 2y/x = 1/(x²e^x)

We must find u such that 2/x u = du/dx:
2/x u = du/dx
1/u du = 2/x dx
∫1/u du = ∫2/x dx
ln u = 2 ln x
u=x²

So:

u dy/dx + 2/x u y = u/(x²e^x)
u dy/dx + du/dx y = u/(x²e^x)
d(uy)/dx = u/(x²e^x)
uy = ∫u/(x²e^x) dx = ∫e^(-x) dx = -e^(-x)+C
x²y = -e^(-x) + C
y=-1/(x²e^x) + C/(x²)

Answer 2

divide through by x

dy/dx+(2/x)y=e^(-x)/x^2

here we will use the inegrating factor method

DE's of the form dy/dx+g(x)y=h(x)

g(x)=2/x,h(x)=e^(-x)/x^2

a) IF=e^(int(2/x))=e^(ln(x^2))
=x^2

b)rewrite DE as

d((x^2)y/dy =(x^2)(e^(-x))/(x^2)
=e^(-x)

c) (x^2)y = int(e^(-x))
= -e^(-x) +C

d) general solution is therefore

y= (-e^(-x)+C)/x^2

i hope that this helps

Answer 3

dividing by x gives you

dy/dx+(2/x)y=(e^-x) / (x^2)

which is a linear equation with P = 2/x and Q=(e^-x) / (x^2)

integrating factor
I.F = e^(integral P dx) = e^(2 ln(x)) = e^(ln ( x^2) ) = (x^2)


solution is y(I.F) = integral {Q* IFdx}

substituting and simplifying,
y*(x^2) = integral{ (e^-x) dx}

or y*(x^2) = -(e^-x) +C

Answer 4

eleventy seven minus the square of the left dangle plus three monkeys equal the same as a left plop