# Show that #y=2/3e^x+e^(-2x)# is a solution of the differential equation #y'+2y=2e^x#?

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I am having trouble coming up with a method to the solution of this general differential equation without the use of an integration factor. It is the first question in Ch. 9.1 of James Stewart's *Calculus*. At this point, the integration factor has not been introduced.

Thanks!

I am having trouble coming up with a method to the solution of this general differential equation without the use of an integration factor. It is the first question in Ch. 9.1 of James Stewart's *Calculus*. At this point, the integration factor has not been introduced.

Thanks!

##### 3 Answers

I don't have that text to refer to, so I will try but depending upon how far you have progressed in solving DE's it may be a out of scope

#y'+2y=2e^x#

This is a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

# dy/dx + P(x)y=Q(x) #

If we had knowledge of the Integrating Factor then we could use:

#IF=e^(int P(x) dx)#

In this example

We could just take the answer and take its derivative and show that it satisfies the differential equation but that is not a method of "coming up with a method to the solution", so let's assume we have no knowledge of the solution.

There are two methods I could use to solve this equation.

**Method 1 - Use the fact that it is a linear DE with constant coefficients**

We use the same method used to solve a second order (or in fact any order) differential equation with constant coefficients. We can solve the homogeneous equation to find the Complementary Function (CF):

#y'+2y = 0#

by turning to the Auxiliary Equation. This is made up with the coefficients of the derivatives, viz;

#m+2=0 => m=-2#

And so we can immediately conclude that

We now need to find a particular integral (PI) solution, so because

#y=alphae^x => y'=alphae^x#

Substitute this into the DE, and we get

# y'+2y = {alphae^x} + 2{alphae^x}#

# " "= alphae^x + 2alphae^x#

# " "= 3alphae^x#

We require that

Hence the general solution is :

#y = CF + PI #

# \ \ = Ae^(-2x) + 2/3e^x # , consistent with the given solution

**Method 2 -Use an Integrating Factor but from First Principles**

#y'+2y=2e^x " "# ,or#dy/dx+2y=2e^x#

We cannot separate the variables for this DE because we cannot find:

#int (dy/dx+2y) dy#

The trick is to find a way to manipulate

If we multiplied the solution

#d/dx(mu(x)y(x)) = mu(x)dy/dx+(dmu)/dxy(x)#

If we multiply both sides of our DE by

#dy/dx+2y=2^x => mu(x)dy/dx+2mu(x)y=2mu(x)e^x#

By comparing our two expressions then we can see that we should choose

#(dmu)/dx=2mu#

Which we can solve as it is a First Order separable DE so we can find the

# int \ 1/mu \ dmu=int \ 2 \ dx#

# :. ln mu = 2x#

# :. mu = e^(2x)#

So if we now multiply our DE by this factor

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

# :. d/dx{e^(2x)y} = 2e^(2x)e^x#

# :. d/dx{e^(2x)y} = 2e^(3x)#

Which is now a separable DE which we can just integrate to get:

# e^(2x)y = int \ 2e^(3x) \ dx #

# :. e^(2x)y = 2/3e^(3x) + A " "# , where#A# is an Arbitrary constant.

# :. y = 2/3e^(3x)e^(-2x) + Ae^(-2x) #

# :. y = 2/3e^x + Ae^(-2x) # , as above.

If you are asked to SHOW that this is a solution, you do not need to solve the DE.

#### Explanation:

Differentiate

Multiply

Add

**Note**

I do not have a copy of Stewart here, but if this is the first exercise in the first section on differential equations, then the point may be to show that this is a solution, not to show how to arrive at this solution.

In a similar way, we might ask an algebra student to show that

See explanation.

#### Explanation:

Showing that something is a solution to an equation means plugging in the given expression (and any implied expressions necessary) to that equation, and making sure it holds true.

Here, we are given the differential equation

Since the differential equation needs

#color(white)(=>')y=2/3 e^x + e^(-2x)#

#=>y'=dy/dx#

#color(white)(=>y')=2/3 e^x - 2e^(-2x)#

Great—now we have expressions for both

#" "color(blue)(y')" "+" "2color(red)(y)" "=2e^x#

#<=>color(blue)(2/3 e^x - 2e^(-2x))+2[color(red)(2/3 e^x + e^(-2x))]stackrel"? "=2e^x#

#<=>2/3 e^x - cancel(2e^(-2x))+4/3 e^x + cancel(2e^(-2x))stackrel"? "=2e^x#

#<=>" "2 e^x=2e^x#

Since the differential equation holds true for the given