Differential equations?

I've already worked out the following, just need to check I'm doing it right...

Use seperation of variables to solve the following differential equations:

(a) dy/dx = (5-x) / y^2

(b) y^1 = y^2 - (e^3t)(y^2) , y(0) = 1

(c) y^1 + y = 1, y(0) = 2

Update:

correction to b and c...

(b) y' = y^2 - (e^3t)(y^2) , y(0) = 1

(c) y' + y = 1, y(0) = 2

y^2 is y squared

2 Answers

Relevance
  • 9 years ago
    Favourite answer

    See:

    (a) dy/dx = (5-x) / y^2

    y^2.dy= (5-x)dx

    ∫y^2.dy= ∫(5-x)dx

    (y^3)/3=5x-(x^2)/2+c

    y^3=15x-3x^2/2+3c

    y=³√(15x-3x^2/2+3c)

    3c=k

    y=³√(15x-3x^2/2+k)

    or

    y=(15x-3x^2/2+k)^3/2

  • 9 years ago

    what are y^1 and y^2? how do they differ from y' and y'' respectively?

    a)

    dy/dx = (5-x)/y²

    cross-multiply by y² / dx to get

    y² dy = (5-x) dx

    integrate

    (⅓)y³ = (5x - ½x²) + c

    y³ = 3(5x - ½x²) + 3c

    let 3c = C (3c is just a constant so it can be represented by any letter; in this case C)

    y³ = 3(5x - ½x²) + C

    y = ³√ [ 3(5x - ½x²) + C ]

    b)

    y' = y^2 - (e^3t)(y^2) , y(0) = 1

    y' = y² - (e^3t)(y²)

    y' = y² [ 1 - e^3t ]

    multiply by dt/y² both sides

    dy/dt = y² [ 1 - e^3t ]

    1/y² dy = [ 1 - e^3t ] dx

    integrate

    -1/y = t - ⅓ e^3t + c

    1/y = ⅓ e^3t - t - c

    y = 1 / [ ⅓ e^3t - t - c ]

    initial values; y(0) = 1

    1 = 1 / [ ⅓ - c ]

    [ ⅓ - c ] = 1

    c = -⅔

    y = 1 / [ ⅓ e^3t - t + ⅔]

    you can factor out if you want

    by multiplying by 3/3

    y = 3 / [ e^3t - 3t + 2]

    c)

    i assume y is differentiated with respect to x

    y' + y = 1, y(0) = 2

    dy/dx = 1 - y

    dy = (1 - y) dx

    1/(1 - y) dy = dx

    -1/(y - 1)dy = dx

    1/(y - 1)dy = -dx

    integrate

    ln(y - 1) = -x + c

    ln(1 - y) = -x + c

    initial values; y(0) = 2

    ln(2 - 1) = c

    c = ln1

    c = 0

    ln(y - 1) = -x

    y - 1 = e^(-x)

    y = 1 + e^(-x)

Still have questions? Get answers by asking now.