# General solution of ordinary differential equation

• tracedinair
In summary, the general solution of the given differential equation is y = [be^(x(1-λ))] / [e^(x)*(1-λ)] + C.
tracedinair

## Homework Statement

Find the general solution of the differential equation,

y' + y = be^(-λx)

where b is a real number and λ is a positive constant.

## Homework Equations

y' + P(x)y = Q(x)

Integrating factor: e^(∫P(x) dx)

## The Attempt at a Solution

Let P(x) = 1, Q(x) = be^(-λx)

The equation is already in the form y' + P(x)y = Q(x).

So, the integrating fator is I(x) = e^(∫1 dx) = e^(x)

Multiplying both sides by the integrating factor.

e^(x)y + e^(x)y = be^(-λx)e^(x)

(e^(x)y)' = be^(-λx)e^(x)

Now integrating the left hand side,

e^(x)y = be^(-λx)e^(x)

Here is my problem. I don't know where to go from here. How do I integrate the right hand side? That's my main problem.

Any help will be greatly appreciated.

Why not simplify it? e^(-ax) = 1/(e^(ax)).

tracedinair said:

## Homework Statement

Find the general solution of the differential equation,

y' + y = be^(-λx)

where b is a real number and λ is a positive constant.

## Homework Equations

y' + P(x)y = Q(x)

Integrating factor: e^(∫P(x) dx)

## The Attempt at a Solution

Let P(x) = 1, Q(x) = be^(-λx)

The equation is already in the form y' + P(x)y = Q(x).

So, the integrating fator is I(x) = e^(∫1 dx) = e^(x)

Multiplying both sides by the integrating factor.

e^(x)y + e^(x)y = be^(-λx)e^(x)

(e^(x)y)' = be^(-λx)e^(x)

Now integrating the left hand side,

e^(x)y = be^(-λx)e^(x)
Please do not write an equation that you know isn't true! You have integrated the left side but not yet the right side so you know they are not equal. You might write
$$e^xy= \int be^{-\lambda x}e^x dx$$
to indicate that the right has yet to be integrated.

Here is my problem. I don't know where to go from here. How do I integrate the right hand side? That's my main problem.
$$e^{-\lambda x}e^x= e^{(1-\lambda)x}$$
Can you integrate that?

Any help will be greatly appreciated.

I apologize for the integral symbol, simple mistake.

I integrated e^((1-λ)x) and came up with this,

∫e^((1-λ)x) dx

Let u = ((1-λ)x) so du = (1-λ) dx

Now,

∫e^(u) du/(1-λ)

1/(1-λ) ∫e^(u) du

1/(1-λ)*e^(u) + C

Substituting back in for u,

1/(1-λ)*e^(x(1-λ)) + C

Going back to e^(x)y = ∫be^(-λx)e^(x) dx, I obtained,

y = be^(-x)*e^(x(1-λ))/(1-λ)

y = [be^(x(1-λ))] / [e^(x)*(1-λ)]

## 1. What is a general solution of ordinary differential equation?

A general solution of ordinary differential equation is a solution that contains all possible solutions to the given differential equation. It is expressed in terms of one or more arbitrary constants, which can take any value. This solution is not unique and can be used to find specific solutions by assigning values to the arbitrary constants.

## 2. How is a general solution of ordinary differential equation different from a particular solution?

A particular solution is a specific solution to the given differential equation, obtained by assigning values to the arbitrary constants in the general solution. On the other hand, a general solution contains all possible solutions and is not unique. It represents a family of solutions, while a particular solution represents a single solution.

## 3. What are the methods for finding a general solution of ordinary differential equation?

There are several methods for finding a general solution of ordinary differential equation, including separation of variables, integrating factors, and the method of undetermined coefficients. Each method is suitable for different types of differential equations and involves a series of steps to obtain the general solution.

## 4. Can a general solution of ordinary differential equation always be found?

Not all ordinary differential equations have a general solution. Some differential equations do not have an analytical solution, and in these cases, a numerical method must be used to approximate the solution. Additionally, some differential equations may have a general solution that is too complex to be expressed in terms of elementary functions.

## 5. How is a general solution of ordinary differential equation verified?

To verify a general solution of ordinary differential equation, it must be substituted into the original differential equation to see if it satisfies the equation. If the general solution satisfies the equation, it is considered a valid solution. Additionally, the general solution can be graphed to visualize the family of solutions and see if it aligns with the given differential equation and initial conditions.

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