Solve ODE System: x'=x+y², y'=-y

[Buri]

Homework Statement

Solve this system:

x' = x + y²
y' = -y

The Attempt at a Solution

My text solves this by guessing a particular solution. It says:

For the second equation y' = -y yields y(t) = y_0(e^-t). Inserting thisinto the first equation, we must solve:

x' = x + (y_0)²(e^-2t)

This is a first-order nonautonomous equation whose solutions may be determined as in calculus by "guessing" a particaular solution of the form ce^-2t. Inserting this quess into the equation yields a particular solution:

x(t) = (-1/3)(y_0)²e^-2t

Hence any function of the form

x(t) = ce^t - (1/3)(y_0)²e^-2t

is a solution of this equation, as is easily checked.

The general solution is then:

x(t) = (x_0 + (1/3)(y_0)²)e^t - (1/3)(y_0)²e^-2t
y(t) = (y_0)e^-t

I don't see how they got the (x_0 + (1/3)(y_0)²) part? I'm not doing this by guessing but trying to solve this system using integrating factors...but I'm not getting the same answer. Any help?

 

[hunt_mat]

I believe that they are using an integrating factor:
$$\frac{dx}{dt}=x+y_{0}^{2}e^{-2t}$$
So
$$\frac{d}{dx}(e^{-t}x(t))=y_{0}^{2}e^{-3t}$$
Integrate and re-arrange to obtain the answer

 

[HallsofIvy]

Homework Statement

Solve this system:

x' = x + y²
y' = -y

The Attempt at a Solution

My text solves this by guessing a particular solution. It says:

For the second equation y' = -y yields y(t) = y_0(e^-t). Inserting thisinto the first equation, we must solve:

x' = x + (y_0)²(e^-2t)

This is a linear, first order, equation. You can solve it, as hunt_mat says, by using the standard formula for an integrating factor. Since the coefficients of x' and x are constants, you can also use a more general method that can be used for higher order linear equations with constant coefficients.

The "related homogeneous equation" is x'= x (drop the function of t) which has general solution $x= x_0e^t$. Now look for a specific solution to the entire equation. Because derivatives of $e^{at}$ always involve $e^{at}$, to get $e^{-2t}$, we should try something of the form $x= Ae^{-2t}$. Then $x'= -2Ae^{-2t}$ and the equation becomes $-2Ae^{-2t}= Ae^{-2t}+ y_0^2e^{-2t}$. Dividing through by $e^{-2t}$ (which is never 0), we have $-2A= A+ y_0^2$ so that $-3A= y_0^2$ and $A= -y_0^2/3$.

Because the equation is linear, we can construct the general solution to the entire equation by adding the general solution to the related homogeneous equation and any single solution to the entire equation. That is, the general solution to the equation is
$$x= x_0e^t- (y_0^2)/3)e^{-2t}$$
and, of course,
$$y= y_0e^{-t}$$

Here is yet another method of solving a non-homogeneous linear equation that does not involve "guessing" at the $e^{-2t}$ dependence. Having determined that $e^{t}$ is a solution to the related homogeneous equation, look for a solution of the form $x= u(t)e^t$. (This is called "variation of parameters" because we are allowing the constant, C, in the solution to the related homogeneous equation, is allowed to vary and become a function of t.) Then, by the product rule, $x'= u'e^t+ ue^t$ so the equation becomes $u'e^t+ ue^t= ue^t+ y_0^2e^{-2t}$. The two terms $ue^t$ cancel precisely because $e^t$ is a solution to the related homogeneous equation and we are left with $u'e^t= y_0^2e^{-2t}$ so that $u'= y_0^2e^{-3t}$. Integrating, $u(t)= -(y_0^2/3)e^{-3t}+ C$.

Then $y= u(t)e^t= (-(y_0^2/3)e^{-3t}+ C)e^t= -(y_0^2/3)e^{-2t}+ Ce^t$.

These methods for solving linear, non-homogeneous, differential equations are especially important with equations of higher order and you will learn them when you start working on linear equations of order 2, 3, or higher.

This is a first-order nonautonomous equation whose solutions may be determined as in calculus by "guessing" a particaular solution of the form ce^-2t. Inserting this quess into the equation yields a particular solution:

x(t) = (-1/3)(y_0)²e^-2t

Hence any function of the form

x(t) = ce^t - (1/3)(y_0)²e^-2t

is a solution of this equation, as is easily checked.

The general solution is then:

x(t) = (x_0 + (1/3)(y_0)²)e^t - (1/3)(y_0)²e^-2t
y(t) = (y_0)e^-t

I don't see how they got the (x_0 + (1/3)(y_0)²) part? I'm not doing this by guessing but trying to solve this system using integrating factors...but I'm not getting the same answer. Any help?

 

[Buri]

Thank, I realized later that I wasn't doing anything about the intial conditions and that's where the (x_0 + (1/3)(y_0)²) comes from. I've actually finished the course now and we did the variation of parameters and did higher dimensional planar systems so I was familiar with everything you said, but didn't think of trying to solve it another way. Thanks.