# 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

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

- Maria DescartesLv 79 years agoFavourite 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

- King LeoLv 79 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)