solve the following initial value problem?

Question

a) dy/dx+4y=e^-x ,y(0)=4/3

Answer 1

dy/dx + 4y = exp(-x); y(0) = 4/3

First solve dy/dx + 4y = 0
dy/dx = -4y
dy/y = -4dx,
Integrating both side from y = 4/3 to y and from x = 0 to x,
ln y - ln c = -4x
y/c = exp(-4x)
y = c exp(-4x)

Find the particular solution by substituting y = k exp(-x),
dy/dx + 4y = k exp(-x)
d [k exp(-x)] / dx + 4k exp(-x) = exp(-x)
-k exp(-x) + 4k exp(-x) = exp(-x)
-k + 4k = 1
k = 1/3

y = c exp(-4x) + 1/3 exp(-x)
y(0) = 4/3 = c + 1/3
c = 1

y = exp(-4x) + 1/3 exp(-x)

Answer 2

An easy way to solve this differential equation is to use an integration factor. If you multiply both sides by e^(Int4 dx ) = e^(4x), you get

e^4x (dy/dx + 4y) = e^(3x). In the right hand side we have the derivative of y e^(4x), so that

d(y e^(4x)/dx = e^(3x). Integrating both sides, we get

y e^(4x) = 1/3 * e^(3x) + C, C the integration constant.

So, y = 1/3 * e^(-x) + C e^(-4x)

Since y(0) = 4/3, it follows that

4/3 = 1/3 * e^(0) + C e^(0) = 1/3 + C => 4/3 - 1/3 = 1.

So, y = 1/3 * e^(-x) + e^(-4x)

Answer 3

Make this into the m+4=0 form, where m will be the exponent of the general solution in the form of Ce^mx.

m=-4 and y=Ae^-4x (general solution)
Particular solution is y=Be^-x.
Full solution is y=Ae^-4x + Be^-x
Plug in x=0: y(0)=A+B=4/3 (one eq. but 2 unknowns).
To create the second equation, differentiate y into the form of the original equation:
dy/dx+4y=e^-x
-4Ae^(-4x)-Be^-x + 4Ae^(-4x) + 4Be^-x = e^-x
simplify:
3Be^-x=e^-x divide both sides by 3e^-x
B=1/3 plug back into first equation y(0)=A+B=4/3
A+1/3=4/3
A=3/3=1

Full solution is then y(x)=e^(-4x)+1/3*e^-x

Answer 4

The solution is the sum of the general homogeneous solution and any particular non-homogeneous solution.
Homogeneous solution is the solution of:
y' + 4y = 0 with the solution: y = e^(-4x) + C.
Particular non-homogeneous solution is the solution of:
y' + 4y = e^(-x) with the solution y = e^(-x)/3.
Therefore the solution is:
y = e^(-4x) + (e^(-x))/3 + C. The initial condition gives C = -2/3.
Therefore your solution is:
y = e^(-4x) + (e^(-x))/3 - 2/3.

Answer 5

homogeneous solution:
y' + 4y =0
then y=ke^-4x where k is a constant
particular solution:
let's put y=ke^-4x + me^-x where m is a constant
y' + 4y = e^-x gives:
-m e^-x + 4m e^-x =e^-x then m=1/3
y(0)=4/3
then k+1/3=4/3 then k=1
y=e^(4x) + (1/3) e^(-x)