Solving 2nd-Order IVP as System of Equations

• MHB
• karush
In summary, the system of equations $u''+u'+4u=\sin t$ has a homogeneous equation of the form $x_2'= \frac{dx}{dt}=-5x_2$ and a derivative of sin(x) is cos(x).
karush
Gold Member
MHB
$\tiny{2.1.5.1.c}$ source
Change the second-order IVP into a system of equations
$\dfrac{d^2x}{dt^2}+\dfrac{dx}{dt}'+4x=\sin t \quad x(0)=4\quad x'(0)= -3$
ok I presume we can rewrite this as
$u''+u'+4u=\sin t$
Let $x_1=u$ and $x_2=u'$ then $x_1'=x_2$
substituting
$x_2'+x_2+4x=\sin t$
$\begin{array}{lllll} &let &x_1=u &and &x_2=u'\\ &then &x_1'=x_2 &and &x_2'=u'' \end{array}$
so
$\begin{array}{llll} x_1'=x_2\\ x_2'=-x_2-4x_1+\sin t \end{array}$

so far

Last edited:
Since $x_2= x_1$ $x_2'= -5x_2+ sin(t)$. The associated homogeneous equation is $x_2'= \frac{dx}{dt}=-5x_2$ which we can write $\frac{dx_2}{x_2}= -5dt$. Integrating both sides, $ln(x_2)= -5t+ C$. Taking the exponential of both sides, $x_2= e^{-5t+ C}= e^Ce^{-5t}= C'e^{-5t}$.

Since the derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x) we look for a solution to the entire equation of the form $x_2= A sin(x)+ B cos(x)$. Then $x_2'= A cos(x)- B sin(x)$ and the equation becomes
$A cos(x)- B sin(x)= -5(A sin(x)+ B cos(x))+ sin(x)$ so that $(A- B)cos(x)+ (5A- B)sin(x)= sin(x)$.

We must have A- B= 0 and 5A- B= 1.

Country Boy said:
Since $x_2= x_1$ $x_2'= -5x_2+ sin(t)$. The associated homogeneous equation is $x_2'= \frac{dx}{dt}=-5x_2$ which we can write $\frac{dx_2}{x_2}= -5dt$. Integrating both sides, $ln(x_2)= -5t+ C$. Taking the exponential of both sides, $x_2= e^{-5t+ C}= e^Ce^{-5t}= C'e^{-5t}$.

Since the derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x) we look for a solution to the entire equation of the form $x_2= A sin(x)+ B cos(x)$. Then $x_2'= A cos(x)- B sin(x)$ and the equation becomes
$A cos(x)- B sin(x)= -5(A sin(x)+ B cos(x))+ sin(x)$ so that $(A- B)cos(x)+ (5A- B)sin(x)= sin(x)$.

We must have A- B= 0 and 5A- B= 1.
mahalo that helped a lot...

1. What is a 2nd-order IVP?

A 2nd-order IVP (initial value problem) is a type of differential equation that involves a second derivative of a function and also includes initial conditions, which are values of the function and its first derivative at a specific point. The goal is to find a function that satisfies the given equation and initial conditions.

2. Why is it useful to solve a 2nd-order IVP as a system of equations?

Solving a 2nd-order IVP as a system of equations can provide a more efficient and organized way to find a solution. By breaking down the problem into multiple equations, it becomes easier to manipulate and solve for the unknown variables. Additionally, solving as a system allows for the use of linear algebra techniques, which can be helpful in more complex problems.

3. What are the steps to solving a 2nd-order IVP as a system of equations?

The first step is to rewrite the 2nd-order differential equation as a system of two first-order equations by introducing a new variable. Then, use the initial conditions to form a system of equations. Next, solve the system using methods such as substitution or elimination. Finally, use the solution to find the original function.

4. Can you provide an example of solving a 2nd-order IVP as a system of equations?

Sure, let's say we have the 2nd-order IVP: y'' + 4y' + 3y = 0, with initial conditions y(0) = 1 and y'(0) = 2. We can rewrite this as the system of equations: x' = y and y' = -4y - 3x, with initial conditions x(0) = 1 and y(0) = 2. Solving this system, we get x = e^(-t) + 2e^(-3t) and y = e^(-t) - e^(-3t). Therefore, the solution to the original 2nd-order IVP is y = e^(-t) - e^(-3t).

5. What are some applications of solving 2nd-order IVPs as systems of equations?

2nd-order IVPs arise in many fields of science and engineering, such as physics, chemistry, and economics. Solving them as systems of equations can be useful in modeling and predicting systems that involve acceleration, oscillations, or growth and decay. For example, in physics, solving 2nd-order IVPs can help to determine the motion of a pendulum or the displacement of a spring.

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