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### General and Particular Solutions

```
Date: 01/08/99 at 05:57:52
From: Abbas
Subject: Differential Equations

Can you please explain me in simple terms the difference between a
general solution and a particular solution of a differential
equation? When does this occur, and what is its physical significance?
```

```
Date: 01/08/99 at 10:29:49
From: Doctor Rob
Subject: Re: Differential Equations

This is analogous to talking about a general triangle, or a particular
one with sides 7, 8, and 10. The general solution encompasses all
solutions, and a particular solution is just one of those.

It always occurs when you are solving differential equations.

Its physical significance is that the particular solution solves the
equation, but may not satisfy the initial conditions you are interested
in. The solution that does that will be (usually) a different
particular solution of the equation, which you can discover from the
general solution (which depends on some parameters, such as constants
of integration) by applying the initial conditions and solving for the
parameter values which make those conditions true.

Example: Solve the differential equation:

dy/dx = y + 1,    y(0) = 1

The general solution is:

y = c*e^x - 1

for any real constant c. The particular solution you are interested in
is the one where y(0) = 1, that is, c*e^0 - 1 = 1, that is, c = 2, that
is:

y = 2*e^x - 1

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 01/09/99 at 00:28:14
From: Abbas Bookwala
Subject: Re: Differential Equations

Dear Sir,

I am sorry for not forwarding the question with a bit of extra
information. I was in fact interested in knowing those general and
particular solutions occurring in certain equations which are added and
the sum is called a solution. If my memory serves me right (back to
college days), it was a solution of a non-homogeneous equation. I want
to know the physical significance of general and particular solutions
in this context.

Regards,
Abbas S Bookwala
```

```
Date: 01/11/99 at 11:01:10
From: Doctor Rob
Subject: Re: Differential Equations

This new situation you describe is not unrelated to the situation I

When you have a linear differential equation of the form:

a(0,x)*y + a(1,x)*y' + a(2,x)*y" + ... = f(x)

often it is not terribly hard to find the general solution of the
corresponding homogeneous diffential equation:

a(0,x)*y + a(1,x)*y' + a(2,x)*y" + ... = 0

Call that y_0(x,p1,p2,...). It depends on some parameters p1, p2, etc.
Often you can also find a particular solution (depending on no
parameters) y_1(x). Since the equation is linear, then y = y_0 + y_1 is
also a solution of the original equation, for any values of the
parameters. This is pretty easy to see by substituting for y its equal
y_0 + y_1, using the fact that the derivative of a sum is the sum of
the derivatives of the summands, and the fact that y_0 satisfies the
homogeneous equation and that f_1 satisfies the original equation.

As for the physical significance, I suppose you can think of the
homogeneous equation as the one which drives the behavior of the
solutions under some kind of ideal or simplified conditions, and the
presence of the function f(x) is forcing some kind of perturbation to
that behavior. Once you find how that perturbation affects any single
solution (such as y_1), that tells you how it affects all the rest,
because the lefthand side is linear.

The same example works here. Write the equation in the form:

y' - y = 1

The corresponding homgeneous equation is:

y' - y = 0

whose general solution is y_0(x,c) = c*e^x. We can look for a
particular solution which is a polynomial in x, say y = a*x + b, so
y' = a:

a - (a*x + b) = 1

which must be an identity. This forces a = 0, b = -1, and, sure enough,
y_1(x) = -1 is a solution to the inhomogeneous equation. Putting these
together, the general solution to the original equation is:

y(x) = y_0(x,c) + y_1(x)
= c*e^x - 1

Perhaps a more complicated example would be more enlightening:

x*y' - y = x^2

The homogeneous equation is:

x*y' - y = 0
y'/y = 1/x
log|y| = log|x| + C
y = e^C*x
y = D*x

for any value of D. This gives us y_0(x,c) = D*x. Now we want the
particular solution. Again, we try a polynomial of low degree:

y = a*x^2 + b*x + c
y' = 2*a*x + b
x*(2*a*x + b) - (a*x^2 + b*x + c) = x^2
(2*a-a-1)*x^2 + (b-b)*x - c = 0
c = 0,  a = 1,  b = anything

y_1(x) = x^2

Then the general solution is:

y(x) = D*x + x^2

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus

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