Solving a System of Equations: Step-by-Step Guide

In summary, a system of equations is a set of two or more equations with multiple variables that can be solved to find the values of the variables that satisfy all of the equations. This is important in real-world applications for finding solutions to problems with multiple constraints. The most common methods for solving a system of equations are substitution, elimination, and graphing. To solve a system of equations using substitution, you must first solve one equation for a variable, substitute it into the other equation, solve for the remaining variable, and then check the solution. A system of equations can have one solution, infinite solutions, or no solutions depending on the relationship between the equations.
  • #1
squenshl
479
4

Homework Statement


Consider the following system of equations
da/dt = -kab, db/dt = kab, a(0) = a0, b(0) = b0.

Solve these equations exactly.


Homework Equations





The Attempt at a Solution


I added them together to get d(a+b)/dt = 0 which implies a + b = a0 + b0.
Therefore a = a0 + b0 - b so eliminating a we get db/dt = k(a0 + b0 - b)b which is separable, but I don't know where to go from here.

Someone please help.
 
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  • #2
squenshl said:

Homework Statement


Consider the following system of equations
da/dt = -kab, db/dt = kab, a(0) = a0, b(0) = b0.

Solve these equations exactly.


Homework Equations





The Attempt at a Solution


I added them together to get d(a+b)/dt = 0 which implies a + b = a0 + b0.
Therefore a = a0 + b0 - b so eliminating a we get db/dt = k(a0 + b0 - b)b which is separable, but I don't know where to go from here.

Someone please help.
So db/(a0 + b0 - b)b = k
[tex]\frac{db}{b(a_0 + b_0 - b)} = k\cdot dt[/tex]

The left side can be integrated by using partial fraction decomposition. You could simplify the work slightly by rewriting a0 + b0 as, say, M.
 
  • #3
Thanks.

I let M = a0 + b0 and got 1/(b(M-b)) = 1/(Mb) + 1/(M(M-b))
and this integrates to (ln(b)-ln(M-b))/M,
therefore we get (ln(b)-ln(M-b))/M = kt + c, so ln(b/(M-b)) = M(kt+c), so b/(M-b) = exp(M(kt+c) but what do I do now.
 
  • #4
Then b = (M - b)exp(M(kt + c))
==> b - bexp(M(kt + c)) = Mexp(M(kt + c))
==> b(1 - exp(M(kt + c))) = Mexp(M(kt + c))
==> b = ?

You should be able to get rid of the constant c, since you are given that b(0) = b0.

Finally, since a and b add up to a constant, you can solve for a.

When you get a, by all means, check your work. Check that a(0) and b(0) turn out as expected, and then check that a'(t) = -kab, and that b'(t) = -a'(t).
 
  • #5
We get b = Mexp(M(kt+c))/(1-exp(M(kt+c))
so b(0) = b0 = Mexp(Mc)/(1-exp(Mc)), but how do we find c?
 
  • #6
squenshl said:
We get b = Mexp(M(kt+c))/(1-exp(M(kt+c))
so b(0) = b0 = Mexp(Mc)/(1-exp(Mc)), but how do we find c?

b0(1 - exp(Mc)) = Mexp(Mc)
==> b0 - b0exp(Mc) = Mexp(Mc)
==> b0 = b0exp(Mc) + Mexp(Mc) = exp(Mc)(b0 +M)
==> b0/(b0 +M) = exp(Mc)

Now take the ln of both sides.
 

FAQ: Solving a System of Equations: Step-by-Step Guide

What is a system of equations?

A system of equations is a set of two or more equations that contain multiple variables. The solution to a system of equations is the set of values for the variables that satisfy all of the equations simultaneously.

Why do we need to solve a system of equations?

Solving a system of equations allows us to find the values of the variables that satisfy all of the equations. This is useful in real-world applications, such as finding the intersection point of two lines or determining the optimal solution to a problem with multiple constraints.

What are the different methods for solving a system of equations?

The most common methods for solving a system of equations are substitution, elimination, and graphing. Substitution involves solving one equation for a variable and substituting it into the other equation. Elimination involves adding or subtracting the equations to eliminate one of the variables. Graphing involves graphing both equations and finding the point where they intersect.

What is the step-by-step guide for solving a system of equations using substitution?

Step 1: Solve one of the equations for a variable.Step 2: Substitute the solved expression into the other equation.Step 3: Solve for the remaining variable.Step 4: Substitute the value found in Step 3 into either of the original equations to find the value of the other variable.Step 5: Check the solution by substituting the values into both equations and ensuring they both equal.

How do I know if a system of equations has one solution, infinite solutions, or no solutions?

A system of equations has one solution if the equations intersect at one point. It has infinite solutions if the equations are equivalent, meaning they represent the same line. It has no solutions if the equations are parallel and do not intersect.

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