Question

Solve this question (differential equations)
Xd²y/dx²-(2x-1)dy/dx+(x-1)y=0

Answer 1

x^2 dy/dx + xy = 1

I know I have to get all the y's on one side with the dy and the x's on the other with dx, but I can't seem to rearrange this.

my attempt:

x^2dy + xydx = dx
x(xdy + ydx) = dx
xdy + ydx = dx / x
xdy = dx(1/x - y)
xdy/dx=1/x - y

Answer 2

y = (e^x)(c1 log(x) + c2) is the solution to the differential equation xd²y/dx²-(2x-1)dy/dx+(x-1)y=0

Given,

The differential equation, (x) y" - (2x-1) y' + (x-1) y = 0

To Find,

The solution of the differential equation

Solution,

The equation we have been given is a second-order homogenous differential equation with variables.

The general form of a second-order homogenous differential equation is y" + p(x) y' + q(x) y = 0.

Converting the given equation to general form could be done by dividing both the LHS and RHS of the equation by x.

⇒ y" - (2 - 1/x) y' + (1 - 1/x) y = 0

Let this be equation (1)

Here, P(x) = -(2 - 1/x) = 1/x - 2

Q(x) = 1 - 1/x

The solution for this equation could be found using the following steps:

1. Find out the value of 1 + P(x) + Q(x)
⇒ 1 + P(x) + Q(x) = 1 + 1/x - 2 + 1 - 1/x
⇒ 1 + P(x) + Q(x) = 0

2. Since we have the value of 1 + P(x) + Q(x) as zero, we have e^x as a solution to the given equation.

3. To find the complete equation, substitute x=e^x in y=vx

⇒ y = v(e^x)

4. Differentiate the above equation with respect to x:

⇒ y' = v(e^x) + v'(e^x)

⇒ y' = (v + v')(e^x)

Let this be equation (2)

4. Differentiate again with respect to x:

⇒ y" = v(e^x) + v'(e^x) + v'(e^x) + v"(e^x)

⇒ y" = (v + v' + v' + v")(e^x)

⇒ y" = (v"+ 2v' + v)(e^x)

Let this be equation (3)

5. Substitute equations (2) and (3) expression in equation (1):

⇒ (v" + 2v' + v)(e^x) - (2 - 1/x)(v + v')(e^x) + (1 - 1/x) (ve^x) = 0

6. Divide throughout by e^x:

⇒ (v" + 2v' + v) - (2 - 1/x)(v + v') + (1 - 1/x) (v) = 0

⇒ v" + 2v' + v - 2v - 2v' + v/x + v'/x + v - v/x = 0

⇒ v" + v'/x = 0

⇒ v" + (1/x)v' + 0.v = 0

Let this be equation (4)

7. Let v' = p. then v" = p'. Then equation (4) becomes

⇒ p' + (1/x)p = 0

Rearranging this equation, we get

⇒ p' = -(1/x)p

⇒ p' = -(p/x)

⇒ dp/dx = -(p/x)

⇒ dp/p = -(dx/x)

Let this be equation (5)

8. Integrate equation (5):

⇒ ∫(dp/p) = ∫-(dx/x)

⇒ ∫(1/p)dp = -∫(1/x)dx

⇒ log(p) = - log(x) + log(c1)

⇒ log(p) + log(x) = log(c1)

⇒ log(px) = log(c1)

Taking antilog on both sides:

⇒ px = c1

9. We know that p = v'

⇒ v'(x) = c1

⇒ v' = c1/x

⇒ dv/dx = c1/x

Rearranging this equation, we get:

⇒ dv/c1 = dx/x

Let this be equation (6)

10. Integrate equation (6):

⇒ ∫(dv/c1) = ∫(dx/x)

⇒ (1/c1) ∫dv = ∫(1/x)dx

⇒ ∫dv = c1 ∫(1/x)dx

⇒ v = c1 log(x) + c2

11. Substitute, the above value of v in y = v(e^x)

⇒ y = (e^x)(c1 log(x) + c2)

This is the required solution.

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