The problem reads:

The solution of a certain differential equation is of the form

$$y(t)= a\exp(5t) + b\exp(8t)$$

where $a$ and $b$ are constants.

The solution has initial conditions $y(0)=5$ and $y'(0)=5$

Find the solution by using the initial conditions to get linear equations for $a$ and $b$. ....................

What I did was solve using the initial conditions and I found that

$a + b = 5 $

and $ 5a + 8b =5. $

Am I totally on the wrong track? I don't know what it means to find a linear equation for $a$ and $b$. I'd appreciate it if you could solve it step by step.

  • 2
    $\begingroup$ You are on the correct way. You have used the initial condition $y(0)=5$ to get the linear equation $a+b=5.$ And, you have used $y'(0)=5$ to get the linear equation $5a+8b=5.$ Now, solving the system (of linear equations) you get the values of $a$ and $b.$ $\endgroup$ – mfl Sep 4 '16 at 0:09

You're absolutely right.

If $y(t) = a\mathrm e^{5t} + b\mathrm e^{8t}$ then $y'(t) = 5a\mathrm e^{5t} + 8b\mathrm e^{8t}$.

Hence $y(0) = a\mathrm e^0 + b\mathrm e^0 = a+b$ and $y'(0) = 5a\mathrm e^0 + 8b\mathrm e^0 = 5a+8b$.

To solve $y(0)=5$ and $y'(0)=5$ you need to solve $a+b=5$ and $5a+8b=5$ simultaneously.

Personally I would use matrix algebra, but it's up to you.

$$\left(\begin{array}{cc} 1 & 1 \\ 5 & 8 \end{array}\right) \left(\begin{array}{c} a \\ b \end{array}\right)= \left(\begin{array}{c} 5 \\ 5 \end{array}\right) $$

The two-by-two matrix on the right has determinant $1\times 8 - 5\times 1 = 3 \neq 0$ and so there is a unique solution. Multiplying the left and right by the inverse matrix gives

$$ \left(\begin{array}{c} a \\ b \end{array}\right)= \frac{1}{3}\left(\begin{array}{cc} 8 & -1 \\ -5 & 1 \end{array}\right) \left(\begin{array}{c} 5 \\ 5 \end{array}\right) $$

Expanding the right hand side gives \begin{eqnarray*} a &=& \frac{1}{3}(8\times 5 - 1\times 5) &=& \frac{35}{3} \\ \\ b &=& \frac{1}{3}(-5\times 5 + 1 \times 5) &=&-\frac{20}{3} \end{eqnarray*}

Your final solution is then

$$\boxed{y(t) = \tfrac{35}{3}\mathrm e^{5t} - \tfrac{20}{3}\mathrm e^{8t}} $$


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