How do you solve #y'+3y=0# given y(0)=4?

We have # y'+3y=0 #, or:

# dy/dx + 3y = 0 #

This is a first order linear ordinary differential equation and so we can derive an Integrating Factor using:

IF # = e^(int 3dx) #
# \ \ \ = e^(3x) #

So multiplying the DE by the IF gives:

# e^(3x)dy/dx + 3ye^(3x) = 0 #

Thanks to the product rule this can now be written as the derivative of a single product:

# d/dx(ye^(3x)) = 0 #

Which can now easily be solved to give;

# ye^(3x) = A #
# :. y = Ae^(-3x) #

We know that #y(0)=4 => Ae^0=4 => A=4 #

Hence, the solution is:
# y = 4e^(-3x) #

We have the separable differential equation:

#y'+3y=0#

Which can be rearranged as:

#dy/dx+3y=0#

#dy/dx=-3y#

Rearranging this by separating the variables, that is, treating #dy/dx# as division and getting the #y# terms on one side and the #x# terms on the other, we see that:

#dy/y=-3dx#

Integrate both sides:

#intdy/y=-3intdx#

#lnabsy=-3x+C#

Use the initial condition #y(0)=4# to solve for #C#:

#lnabs4=-3(0)+C#

#C=ln(4)#

Then:

#lnabsy=-3x+ln(4)#

Solving for #y#:

#absy=e^(-3x+ln(4))#

Rewriting using exponent rules:

#absy=e^(-3x)e^ln(4)#

#absy=4e^(-3x)#

#4e^(-3x)# is positive for all real values of #x#, so the absolute value bars are not necessary.

#y=4e^(-3x)#