ordinary differential equations
I am asked to solve a second order differential equation to solve with two initial conditions.
$6y'' – 5y' +y = 0$
$y(0)=4$
$y'(0)= 0$
I have the solution which is…
$y = 12e^{\frac{1}{3}x} – 8e^{\frac{1}{2}x}$
But the second portion of the question is asking to sketch the solution. I'm drawing a complete blank on this.
Graphing calculators aren't allowed on tests, so I need to see how it can be done by hand. Thanks.
I'd say it is finite difference method. Approximate the derivative by the central second order difference quotient $$ \frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x)+O(h^2) $$ and apply it to the grid with $x_k=x_0+k·h$, $x_f=x_0+N·h$ over the interval $[x_0,x_f]$.
In case of linear ODE with constant coefficients, it is known that the general solution is a linear combination of exponential functions on the form $k\,e^{r\,t}$ where the various values of $r$ , real or complex, are the roots of a polynomial equation. So, a method to solve the system is to presume of the form of particular solutions and determine the parameters of them, as done below :
Best Answer
You know that $y\to0^+$ as $x\to-\infty$. You know that $y=4$ at $x=0$ and if you solve
$$ 12e^{\frac{1}{3}x} - 8e^{\frac{1}{2}x}=0$$
you get $y=0$ at $x=6\ln\left(\dfrac{3}{2}\right)\approx2.4$.
If you set
$$y^\prime=4e^{\frac{1}{3}x} - 4e^{\frac{1}{2}x}=0$$
you find that $x=0$ so the point $(0,4)$ is actually a maximum point of the graph.
This is enough information to sketch a convincing graph. Be sure to label the coordinates.
In other words try to find asymptotes, intercepts and max/min points using techniques you learned in calculus.