Help With Solving This Differential Equation

[helpmeimdumb]

Homework Statement

1. The slope of a function at any point (x,y) is 〖2e〗^x/(e^x+2) . The point (0,2ln3) is on the graph of f.

(A) Write an equation of the tangent line to the graph of f at x=0.
(B) Use the tangent line in part A to approximate f(0.3) to the nearest thousandth.
(C) Solve the differential equation dy/(dx )=(2e^x)/(e^x+2) with the initial condition f(0)=2ln3.
(D) Use the solution in part C to find f(0.3) to the nearest thousandth.

Homework Equations

The Attempt at a Solution

(A)

dy/(dx )=(2e^x)/(e^x+2)

At x=0, dy/dx=(2e^0)/(e^0+2)=2/3

Equation of tangent line at x=0: y-2ln3=2/3 (x-0)

y-2ln3=2/3 x or y=2/3 x+2ln3

(B)

f(0.3)≈2/3 (0.3)+2ln3≈2.39722≈2.397

(C)

dy/dx=〖2e〗^x/(e^x+2) → dy=(2e^x)/(e^x+2) → ∫dy= ∫〖(2e^x)/(e^x+2) dx〗

Let u=e^x+2, du=e^x dx
2du=2e^x

Here is where I run into problems. Can anybody help me with the rest? Any help would be greatly appreciated.

 

[helpmeimdumb]

bump. please, I'm desperate for help.

 

[SammyS]

Homework Statement

1. The slope of a function at any point (x,y) is 〖2e〗^x/(e^x+2) . The point (0,2ln3) is on the graph of f.

(A) Write an equation of the tangent line to the graph of f at x=0.
(B) Use the tangent line in part A to approximate f(0.3) to the nearest thousandth.
(C) Solve the differential equation dy/(dx )=(2e^x)/(e^x+2) with the initial condition f(0)=2ln3.
(D) Use the solution in part C to find f(0.3) to the nearest thousandth.

Homework Equations

The Attempt at a Solution

(A)

dy/(dx )=(2e^x)/(e^x+2)

At x=0, dy/dx=(2e^0)/(e^0+2)=2/3

Equation of tangent line at x=0: y-2ln3=2/3 (x-0)

y-2ln3=2/3 x or y=2/3 x+2ln3

(B)

f(0.3)≈2/3 (0.3)+2ln3≈2.39722≈2.397

(C)

dy/dx=〖2e〗^x/(e^x+2) → dy=(2e^x)/(e^x+2) → ∫dy= ∫〖(2e^x)/(e^x+2) dx〗

Let u=e^x+2, du=e^x dx
2du=2e^x

Here is where I run into problems. Can anybody help me with the rest? Any help would be greatly appreciated.

Rewrite your integral in terms of the variable, u .