Solve the Differential Equation #dy/dx +3y = 0# with #x=0# when #y=1#?

Question

Isabella Ackerman · Accepted Answer

# y = e^(-x) #

We have: #dy/dx +3y = 0# with #x=0# when #y=1# ..... [A] We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form; # dy/dx + P(x)y=Q(x) # The given equation is already in standard form, So Then the integrating factor is given by; # I = e^(int P(x) dx) # # \ \ = exp(int \ 3 \ dx) # # \ \ = exp( 3x ) # # \ \ = e^(3x) # And if we multiply the DE [A] by this Integrating Factor, #I#, we will have a perfect product differential; # e^(3x)dy/dx +3e^(3x)y = 2e^(-x)e^(3x) # # :. d/dx ( e^(3x)y ) = 2e^(2x) # Which we can directly integrate to get: # e^(3x)y = int \ 2e^(2x) \ dx # # :. e^(3x)y = e^(2x) + C # Using the initial condition #x=0# when #y=1#, we have: # :. e^(0) = e^(0) + C => C = 0 # Thus, # e^(3x)y = e^(2x) # # :. y = e^(2x)e^(-3x) # # \ \ \ \ \ \ \ = e^(-x) # And the given answer is incorrect.

Evelyn Agard · Answer

The solution is:

#y(x) = e^(-x)# Solve the homogeneous differential equation: #dy/dx +3y = 0# The characteristic equation is: #lambda + 3 = 0# so the general solution of the homogeneous equation is: #y_0(x) = Ce^(-3x)# Search now a particular solution using Lagrange methods of variable coefficients in the form: #y_1(x) = a(x) bary_0(x) # where #bary_0(x)# is a non trivial solution of the homogeneous equation, and we choose the one for #C=1# so that: #y_1(x) = a(x) e^(-3x)# Substituting into the equation: #(dy_1)/dx +3y_1 =2e^(-x)# #d/dx( a(x) e^(-3x) ) +3 a(x) e^(-3x) = 2e^(-x)# using the product rule: #(da)/dx e^(-3x) -3 a(x) e^(-3x) + 3 a(x) e^(-3x) = 2e^(-x)# #(da)/dx e^(-3x) = 2e^(-x)# #(da)/dx = 2e^(2x)# #a(x) = int 2e^(2x)dx = e^(2x) + c# We can take the solution for #c=0# and have: #y_1(x) = a(x)e^(-3x) = e^(2x)e^(-3x) = e^(-x)# The general solution of the equation is then: #y(x) = Ce^(-3x)+ e^-x# Imposing #y(0) = 1# we get #C=0# and the required solution is: #bary(x) = e^-x# In fact: #d/dx (e^(-x))+3e^(-x) = -e^(-x) +3e^(-x) = 2e^(-x)#

Evelyn Agard · Answer

#y=e^(-3x)#

I try to find solution of #dy/dx+3y=0# differential equation with condition #y(0)=1#

After taking Laplace transformation both sides,

#sY(s)-y(0)+3Y(s)=0#

#(s+3)Y(s)-1=0#

#Y(s)=1/(s+3)#

After taking inverse Laplace transform, I found

#y=e^(-3x)#

Samantha Adkison · Answer

undefined To solve the differential equation $ \frac{dy}{dx} + 3y = 0 $ with the initial condition $ y(0) = 1 $, we can separate variables and integrate both sides.

Starting with the given differential equation:
$$ \frac{dy}{dx} + 3y = 0 $$

Separating variables:
$$ \frac{dy}{y} = -3dx $$

Integrating both sides:
$$ \int \frac{1}{y} dy = \int -3dx $$

$$ \ln|y| = -3x + C $$

where $ C $ is the constant of integration.

Applying the initial condition $ y(0) = 1 $:
$$ \ln|1| = -3(0) + C $$
$$ 0 = C $$

Thus, the particular solution is:
$$ \ln|y| = -3x $$

To find $ y $:
$$ |y| = e^{-3x} $$

Applying the initial condition again:
$$ |1| = e^{0} $$
$$ 1 = 1 $$

So, the solution to the differential equation with the initial condition is:
$$ y(x) = e^{-3x} $$