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

Question

Emily Ammerman · Accepted Answer

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

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) #

Christopher Allers · Answer

#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)#

Adam Adkins · Answer

undefined To solve the differential equation $ y' + 3y = 0 $ with the initial condition $ y(0) = 4 $, you can use the method of separation of variables.

1. Rewrite the equation in the form $ y' = -3y $.
2. Separate variables by dividing both sides by $ y $ and multiplying both sides by $ dx $: $ \frac{dy}{y} = -3dx $.
3. Integrate both sides:
   - Integrate $ \frac{dy}{y} $ to get $ \ln|y| $.
   - Integrate $ -3dx $ to get $ -3x + C $, where $ C $ is the constant of integration.
4. Combine the results: $ \ln|y| = -3x + C $.
5. Solve for $ y $ by exponentiating both sides: $ |y| = e^{-3x + C} $.
6. Apply the initial condition $ y(0) = 4 $:
   - When $ x = 0 $, $ |y| = e^C $.
   - Since $ |y| = 4 $, $ e^C = 4 $.
   - Therefore, $ C = \ln(4) $.
7. Substitute $ C = \ln(4) $ back into the equation: $ |y| = e^{-3x + \ln(4)} $.
8. Simplify and solve for $ y $:
   - $ |y| = e^{\ln(4) - 3x} = 4e^{-3x} $.
   - $ y = \pm 4e^{-3x} $.
9. Apply the initial condition to find the particular solution:
   - $ y(0) = 4 $ implies $ y = 4e^0 = 4 $.
10. Hence, the solution to the initial value problem is $ y = 4e^{-3x} $.