Differential Equations?

Question

Can you help me with this: Solve the differential eq:

2(dy/dx) - 4xy = 5x ; (y=10 when x=1)

This is what i have so far:

dy/dx - 2xy = 5/2x

so, e^(-x^2)(dy/dx) - 2xye^(-x^2) = 5/2xe^(-x^2)

Dx (e^(-x^2)y) = 5/2xe^(-x^2)

I'm stuck. I know the next step involves substitution but I don't know how to proceed.

Thanks for your help everyone! Kathleen: Sure I'd go to a Prof., but as you can see, it's 8:30pm and I don't know any profs that make themselves available this late.

Answer 1

integrate both sides...

e^(-x^2)y = -5/4e^(-x^2) + C

y = -5/4 + Ce^(x^2)

then...
plug in the point (1,10)

10 = -1.25 + Ce
11.25/e = C


thus
y = -5/4 + (11.25/e) e^(x^2)

or

y = -5/4 + (45/4) e^(x^2-1)

§

Answer 2

Take the equation without right side
dy/dx-2xy=0
dy/y= 2xdx so ln I y I = x^2+C
y=C*e^x^2
Find a particular solution
By inspection y=k ( constant ) can be
-4kx=5x so k = -5/4
So the general solution is
y= Ce^x^2 -5/4

10 = C e -5/4 so C = 45/4e and y = 45/4 * e^(x^2-1)-5/4

Answer 3

To Kathleen: Don't have to be in college to do differential equations lol.

2(dy/dx) - 4xy = 5x

dy/dx - 2xy = (5/2)x

dy/dx = (5/2)x + 2xy

dy/dx = x(5/2 + 2y)

dy / (5/2+2y) = x dx

Integral dy / (5/2+2y) = (1/2)x^2 + C

1/2(Integral du / u) = 1/2x^2 + C

1/2(ln |u|) = 1/2x^2 + C

1/2 [ln (5/2 + 2y)] = (1/2)x^2 + C

ln (5/2+2y) = x^2 + C

5/2 + 2y = e^(x^2 + C)

y = [e^(x^2 + C) / 2] + 5/4

10 = [e^(1+C) / 2] + 5/4

35/4 = e^(1+C) / 2

35/2 = e^(1+C)

17.5 = e^(1+C)

ln 17.5 = 1 + C

2.8622... = 1 + C

C = 1.8622...

y = [e^(x^2+1.8622...) / 2] + 5/4

Answer 4

(e^-(x²))y = (5/2) ∫xe^(-x²)dx

∫xe^(-x²)dx
Let u = x²
du = 2xdx
∫xe^(-x²)dx
= ∫(1/2) (e^-u)du
= -1/2 e^-u
= -1/2 e^(-x²) + c


(e^-(x²))y = (5/2) ∫xe^(-x²)dx
(e^-(x²))y = (-5/4)e^(-x²) + c
y = -(5/4) + ce^(x²)

Check:
y' = 0 + 2cxe^(x²)

2y' - 4xy
= 4cxe^(x²) + 5x - 4cxe^(x²)
= 5x
OK

Apply boundary conditions:
y = -(5/4) + ce^(x²)
x=1, y = 10
10 = -5/4 + ce^1
ce = 45/4
c = 45/(4exp(1))

y = -(5/4) + (45/(4exp(1)))e^(x²)