Solving Linear Differential Equations

बनाया गया: 21 नवंबर 2024

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प्रश्न

Here is the extracted text from the image:

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The system of first-order differential equations:
\[
\begin{cases}
y_1' = -13 y_1 + 2 y_2, & y_1(0) = 2 \\
y_2' = -34 y_1 + 3 y_2, & y_2(0) = 13
\end{cases}
\]

has solution:
\[
y_1(t) = \, \text{[ ]}
\]
\[
y_2(t) = \, \text{[ ]}
\]

**Note:** You must express the answer in terms of real numbers only.

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Solve the question by typing the followings: The solution contains atleast two steps. Write the subject or sub-subject. In step 1: What is given, what do we need to find out, explanation part (Define/concept used, formula used), Step-by-step solution. In other steps: Continue the step-by-step solution by giving heading what now we are going to calculate in this step. (In brief)
Final answer in sentence.

विचार

Setting up the system

I’m piecing together a system of first-order linear differential equations, noting the initial conditions. This involves expressing the solutions solely in terms of real numbers.

Piecing together solution format

I’m analyzing the solution instructions, which require clarity, definition, and a structured step-by-step approach, including a heading "Subject or sub-subject" and specifics in each step.

Formulating the system

I am setting up the system in matrix form to simplify solving linear differential equations by pinpointing eigenvalues and eigenvectors.

Calculating eigenvalues

First, I calculated the determinant, set it to zero, and found the eigenvalues. Then, I computed the eigenvectors for each eigenvalue and formulated the general solution.

Mapping the journey

OK, let me see. I’m outlining the process of solving a system of first-order linear differential equations with eigenvalues and eigenvectors. This involves understanding the problem, applying formulas, and detailing the solution step-by-step.

Applying eigenvalues and eigenvectors

OK, let me see. The problem involves solving a system of first-order linear differential equations using eigenvalues and eigenvectors, with initial conditions given. This will allow us to determine functions h1(t)h_1(t) and h2(t)h_2(t).

Structuring the system

I’m piecing together the system into matrix form with y\mathbf{y} and AA to find eigenvalues and eigenvectors for the general solution.

Crunching the numbers

I’m working out the determinant of AλIA - \lambda I to find eigenvalues. Expanding and combining terms, it’s interesting to see the matrix's properties unfold.

Revisiting the equation

I’m working through the product (13λ)(3λ)( -13 - \lambda )( 3 - \lambda ). Initially, I calculated 39+16λλ2-39 + 16 \lambda - \lambda^2, but I’m careful to ensure accuracy with each step.

Crunching the numbers

I expanded (13λ)(3λ)(-13 - \lambda)(3 - \lambda) and added 68. This gives a characteristic equation for eigenvalues.

Evaluating eigenvalues

I determined that the eigenvalues of matrix AA are 5±2i-5 \pm 2i by solving the quadratic equation derived from det(AλI)=0\det(A - \lambda I) = 0.

Breaking down the matrix

I’m subtracting λI\lambda I from matrix AA to get the matrix AλIA - \lambda I, which is (82i23482i)\begin{pmatrix} -8 - 2i & 2 \\ -34 & 8 - 2i \end{pmatrix}.

Solving for v2v_2 in terms of v1v_1

I’m simplifying the first equation to solve for v2v_2 as (4+i)v1(4 + i) v_1. Substituting this into the second equation, I verify that the left side matches the right side, confirming the relationship.

Confirming the equation

Interestingly enough, establishing that v2=(4+i)v1v_2 = (4 + i) v_1 aligns with the equation 34v1=34v134 v_1 = 34 v_1.

Constructing the general solution

I’m working on building the general solution by combining real and imaginary parts of eigenvectors, forming y(t)=eαt[C1cos(βt)+C2sin(βt)]\mathbf{y}(t) = e^{\alpha t} [ \mathbf{C_1} \cos(\beta t) + \mathbf{C_2} \sin(\beta t) ].

Adjusting my approach

I’m piecing together the general solution, ensuring α=5\alpha = -5 and β=2\beta = 2, to find y(t)\mathbf{y}(t).

Mapping out the solution

I'm thinking about expressing the solution for each component, considering the constants c1c_1 and c2c_2. This involves trigonometric terms and vector constants to ensure precision.

Formulating the solution

I’m incorporating the vectors p\mathbf{p} and q\mathbf{q} into the general solution, y(t)\mathbf{y}(t), by combining constants and adjusting for initial conditions.

Weighing the options

I’m mapping out whether to use y(t)=e5t[pcos(2t)qsin(2t)]\mathbf{y}(t) = e^{-5 t} [ \mathbf{p} \cos(2 t) - \mathbf{q} \sin(2 t) ] or explore y1(t)y_1(t) and y2(t)y_2(t) directly.

Rewriting the solution

I’m thinking through a new way to express y(t)\mathbf{y}(t) using constants A\mathbf{A} and B\mathbf{B}, and providing simpler forms for y1(t)y_1(t) and y2(t)y_2(t) with cosine and sine functions, while noting their interdependence.

Expressing the general solution

I’m working through expressing y2(t)y_2(t) in terms of y1(t)y_1(t) using standard theory for systems with complex eigenvalues and eigenvectors. This involves exponential functions and trigonometric terms.

Breaking down the solution

I’m combining constants into c1c_1 and c2c_2, then expanding y(t)\mathbf{y}(t) component-wise to see how p\mathbf{p} and q\mathbf{q} combine with sines and cosines.

Breaking down the processes

I’m decomposing the expressions for y1(t)y_1(t) and y2(t)y_2(t), combining exponential and trigonometric elements to streamline and group terms effectively.

Setting initial values

I determined y1(0)y_1(0) and y2(0)y_2(0), finding c1=2c_1 = 2. Now, I'll solve y2(0)y_2(0) to pinpoint c2c_2.

Breaking down the equation

First, I substitute c1=2c_1 = 2 into y2(0)=4c1+c2=13y_2(0) = 4 c_1 + c_2 = 13, arriving at c2=5c_2 = 5. This leads me to values for c1c_1 and c2c_2, which I'll use to formulate the solutions.

Breaking down the solutions

I derived two distinct solutions for y1(t)y_1(t) and y2(t)y_2(t) by substituting specific values and simplifying each term. The results included exponential decay modulated by trigonometric functions.

उत्तर

Subject: Solving a System of First-Order Linear Differential Equations Using Eigenvalues and Eigenvectors


Step 1: Understanding the Problem

Given:

We have the system of differential equations:

{y1=13y1+2y2,y1(0)=2,y2=34y1+3y2,y2(0)=13.\begin{cases} y_1' = -13 y_1 + 2 y_2, & y_1(0) = 2, \\ y_2' = -34 y_1 + 3 y_2, & y_2(0) = 13. \end{cases}

We need to find: The explicit functions y1(t)y_1(t) and y2(t)y_2(t) that satisfy both the differential equations and the initial conditions.

Explanation:

To solve this system, we'll use the method of eigenvalues and eigenvectors. By expressing the system in matrix form and finding the eigenvalues and eigenvectors of the coefficient matrix, we can construct the general solution.

Step-by-Step Solution:

  1. Write the System in Matrix Form:

Let y=(y1y2)\mathbf{y} = \begin{pmatrix} y_1 \\ y_2 \end{pmatrix}. Then the system becomes:

y=Ay,\mathbf{y}' = A\mathbf{y},

where A=(132343)A = \begin{pmatrix} -13 & 2 \\ -34 & 3 \end{pmatrix}.

  1. Find the Eigenvalues of Matrix AA:

Compute the characteristic equation det(AλI)=0\det(A - \lambda I) = 0:

det((13λ2343λ))=(13λ)(3λ)(34)(2)=[(13λ)(3λ)]+68=0.\begin{aligned} \det\left( \begin{pmatrix} -13 - \lambda & 2 \\ -34 & 3 - \lambda \end{pmatrix} \right) &= (-13 - \lambda)(3 - \lambda) - (-34)(2) \\ &= [(-13 - \lambda)(3 - \lambda)] + 68 = 0. \end{aligned}

Expanding:

(13λ)(3λ)+68=(λ2+10λ+29)=0.(-13 - \lambda)(3 - \lambda) + 68 = (\lambda^2 + 10\lambda + 29) = 0.

Solving the quadratic equation:

λ=10±1001162=5±2i.\lambda = \frac{-10 \pm \sqrt{100 - 116}}{2} = -5 \pm 2i.
  1. Find the Corresponding Eigenvectors:

For λ=5+2i\lambda = -5 + 2i, solve (AλI)v=0(A - \lambda I)\mathbf{v} = 0:

((13(5+2i)2343(5+2i)))v=0,\left( \begin{pmatrix} -13 - (-5 + 2i) & 2 \\ -34 & 3 - (-5 + 2i) \end{pmatrix} \right) \mathbf{v} = \mathbf{0},

which simplifies to:

((82i23482i))v=0.\left( \begin{pmatrix} -8 - 2i & 2 \\ -34 & 8 - 2i \end{pmatrix} \right) \mathbf{v} = \mathbf{0}.

Let v=(v1v2)\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}. From the first row:

(82i)v1+2v2=0    v2=(8+2i)2v1=(4+i)v1.(-8 - 2i)v_1 + 2v_2 = 0 \implies v_2 = \frac{(8 + 2i)}{2}v_1 = (4 + i)v_1.

Thus, the eigenvector is v=v1(14+i)\mathbf{v} = v_1 \begin{pmatrix} 1 \\ 4 + i \end{pmatrix}.

Step 2: Constructing the General Solution

Explanation:

With complex eigenvalues and eigenvectors, the solution involves exponential, sine, and cosine functions.

Step-by-Step Solution:

  1. Express the Eigenvector in Terms of Real and Imaginary Parts:

Separate v\mathbf{v} into real and imaginary parts:

p=(14),q=(01).\mathbf{p} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}, \quad \mathbf{q} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}.
  1. Write the General Solution:

The general solution is:

y(t)=e5t[pcos(2t)qsin(2t)]c1+e5t[psin(2t)+qcos(2t)]c2.\mathbf{y}(t) = e^{-5t} \left[ \mathbf{p} \cos(2t) - \mathbf{q} \sin(2t) \right] c_1 + e^{-5t} \left[ \mathbf{p} \sin(2t) + \mathbf{q} \cos(2t) \right] c_2.
  1. Express y1(t)y_1(t) and y2(t)y_2(t) Separately:
  • For y1(t)y_1(t): y1(t)=e5t[c1cos(2t)+c2sin(2t)].y_1(t) = e^{-5t} \left[ c_1 \cos(2t) + c_2 \sin(2t) \right].
  • For y2(t)y_2(t): y2(t)=e5t[(4c1+c2)cos(2t)+(4c2c1)sin(2t)].y_2(t) = e^{-5t} \left[ (4 c_1 + c_2) \cos(2t) + (4 c_2 - c_1) \sin(2t) \right].
  1. Apply Initial Conditions to Solve for Constants:

At t=0t = 0:

  • y1(0)=c1=2y_1(0) = c_1 = 2.
  • y2(0)=4c1+c2=13y_2(0) = 4 c_1 + c_2 = 13.

Solving:

c1=2,4(2)+c2=13    c2=5.c_1 = 2, \quad 4(2) + c_2 = 13 \implies c_2 = 5.
  1. Write the Final Expressions for y1(t)y_1(t) and y2(t)y_2(t):
  • y1(t)=e5t[2cos(2t)+5sin(2t)]y_1(t) = e^{-5t} \left[ 2 \cos(2t) + 5 \sin(2t) \right].
  • y2(t)=e5t[13cos(2t)+18sin(2t)]y_2(t) = e^{-5t} \left[ 13 \cos(2t) + 18 \sin(2t) \right].

Conclusion:

Therefore, the solutions to the differential equations are:

y1(t)=e5t[2cos(2t)+5sin(2t)],y2(t)=e5t[13cos(2t)+18sin(2t)].\begin{aligned} y_1(t) &= e^{-5t} \left[ 2 \cos(2t) + 5 \sin(2t) \right], \\ y_2(t) &= e^{-5t} \left[ 13 \cos(2t) + 18 \sin(2t) \right]. \end{aligned}

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