Solve the differential equation dy/dx = (1+y)^3?

Question

Please show me how to do these!!!

Answer in my math book is

y=1 + or - (-2(x+c)) ^ -1/2

Answer 1

dy/dx = (1 + y)^3

This is a separable equation. Bring all y terms with the dy, and x terms with the dx.

Multiplying both sides by dx,

dy = (1 + y)^3 dx

Multiplying both sides by 1/(1 + y)^3, we get

1/(1 + y)^3 dy = dx

Which is the same as

(1 + y)^(-3) dy = dx

Taking the integral of both sides,

Integral ( (1 + y)^(-3) dy) = Integral (1 dx )

And now, since (1 + y) is a linear term with y's coefficient 1, we can use the reverse power rule to integrate.

(-1/2)(1 + y)^(-2) = x + C

Answer 2

dy/dx − y³sin(1/x) + (1/x)y = 0 Divide by y³ and express as (1/y³)(dy/dx) + (1/x)(1/y²) = sin(1/x) … (i) Just a Bernoulli form so put z=1/y² with dz/dx=(−2/y³)(dy/dx) (i) becomes dz/dx + (−2/x)z = −2sin(1/x) … (ii) The IF for this is exp( ∫{−2/x}dx ) = exp(−2log(x)) = x‾² Applying the IF to (ii) yields (1/x²)(dz/dx) + (−2/x³)z = 2(−1/x²)sin(1/x) Integrate wrt x, noting that the LHS is always IF*y … z/x² = 2 ∫ (−1/x²)sin(1/x)dx Evaluate the integral using u=1/x to get ∫ (−1/x²)sin(1/x)dx = −cos(1/x) + constant ∴ z/x² = c − 2cos(1/x) Restoring y gives general solution as 1/(x²y²) = c − 2cos(1/x)

Answer 3

dy/dx = (1+y)^3
dy/dx = (1+y)(1+y)(1+y)
dy/dx = (1+y)(1+y)(1+y) divide both sides by "1/dy" or multiply both sides by dx/1

1/dx = (1+y)/dy (1+y)/dy (1+y)/dy
1/dx= 1/d^3 mulitply d/1 by both sides
1/x = 1/d^2

Last time i did this was 5 years ago. I think im wrong. Ill wait to see what other people say cause im curious now

Answer 4

dy/dx = (1+y)³
dy/(1+y)³ = dx
integrating both sides
∫ dy/(1+y)³ = ∫ dx
-1/2(1+y)² = x + c
-1/(1+y)² = 2x +c

Answer 5

∫ dy / (1 + y) ³ = ∫ dx
∫ (1 + y)^(- 3) dy = ∫ dx
(1 + y)^(- 2) / (- 2) = x + C
(1 + y)^(- 2) = - 2x + K
1 / (1 + y) ² = - 2x + K

Answer 6

dy/dx =(1+y)^3
dy/(y+1)^3=dx
(1+y)^-3 dy =dx
integrating both sides

((1+y)-2)/(-2)=x+c

(1+y)^-2=-2x-2c
1=(-2x-2c)(1+y)^2

the term 2c can be simplified as c because a constant times a constant is also constant

Answer 7

dy/dx = 3*(1+y)^2

I trust the question is written currently in that it's dy/dx not dx/dy.

Answer 8

you cant solve this... you'll have an unknown constant still hanging around after u integrate the equation. You are missing out more information.

Answer 9

dy'/dx'=(3+3y)^2

to do these you use the formula ax^n = nax^n-1