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Linear Systems of Differential Equations Calculator

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Example
Created on 2024-06-20Asked by Olivia Baker (Solvelet student)
Solve the system of linear differential equations: {dxdt=3x+4ydydt=−4x+3y \begin{cases} \frac{dx}{dt} = 3x + 4y \\ \frac{dy}{dt} = -4x + 3y \end{cases}

Solution

To solve the system of linear differential equations: {dxdt=3x+4ydydt=−4x+3y \begin{cases} \frac{dx}{dt} = 3x + 4y \\ \frac{dy}{dt} = -4x + 3y \end{cases} we write the system in matrix form: ddt(xy)=(34−43)(xy) \frac{d}{dt} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 & 4 \\ -4 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} The coefficient matrix is: A=(34−43) A = \begin{pmatrix} 3 & 4 \\ -4 & 3 \end{pmatrix} To find the general solution, we need the eigenvalues and eigenvectors of A A . Find the characteristic equation: det⁡(A−λI)=0 \det(A - \lambda I) = 0 det⁡(3−λ4−43−λ)=0 \det \begin{pmatrix} 3 - \lambda & 4 \\ -4 & 3 - \lambda \end{pmatrix} = 0 (3−λ)2−(4)(−4)=0 (3 - \lambda)^2 - (4)(-4) = 0 (3−λ)2+16=0 (3 - \lambda)^2 + 16 = 0 λ2−6λ+25=0 \lambda^2 - 6\lambda + 25 = 0 Solve the characteristic equation: λ=6±36−1002 \lambda = \frac{6 \pm \sqrt{36 - 100}}{2} λ=6±−642 \lambda = \frac{6 \pm \sqrt{-64}}{2} λ=3±4i \lambda = 3 \pm 4i The eigenvalues are λ1=3+4i \lambda_1 = 3 + 4i and λ2=3−4i \lambda_2 = 3 - 4i . Next, find the eigenvectors. For λ=3+4i \lambda = 3 + 4i : (A−(3+4i)I)v=0 (A - (3 + 4i)I)\mathbf{v} = 0 (3−(3+4i)4−43−(3+4i))v=0 \begin{pmatrix} 3 - (3 + 4i) & 4 \\ -4 & 3 - (3 + 4i) \end{pmatrix} \mathbf{v} = 0 (−4i4−4−4i)v=0 \begin{pmatrix} -4i & 4 \\ -4 & -4i \end{pmatrix} \mathbf{v} = 0 Solve for v \mathbf{v} : −4iv1+4v2=0⇒v2=iv1 -4iv_1 + 4v_2 = 0 \Rightarrow v_2 = iv_1 Thus, one eigenvector is v1=(1i) \mathbf{v}_1 = \begin{pmatrix} 1 \\ i \end{pmatrix} . For λ=3−4i \lambda = 3 - 4i , the eigenvector is v2=(1−i) \mathbf{v}_2 = \begin{pmatrix} 1 \\ -i \end{pmatrix} . The general solution is: (xy)=c1e(3+4i)t(1i)+c2e(3−4i)t(1−i) \begin{pmatrix} x \\ y \end{pmatrix} = c_1 e^{(3 + 4i)t} \begin{pmatrix} 1 \\ i \end{pmatrix} + c_2 e^{(3 - 4i)t} \begin{pmatrix} 1 \\ -i \end{pmatrix} Express the solution in real form using Euler's formula: x(t)=e3t[c1cos⁡(4t)+c2sin⁡(4t)]y(t)=e3t[c1sin⁡(4t)−c2cos⁡(4t)] \begin{aligned} x(t) &= e^{3t} \left[ c_1 \cos(4t) + c_2 \sin(4t) \right] \\ y(t) &= e^{3t} \left[ c_1 \sin(4t) - c_2 \cos(4t) \right] \end{aligned} Therefore, the solution is: x(t)=e3t[c1cos⁡(4t)+c2sin⁡(4t)] x(t) = e^{3t} \left[ c_1 \cos(4t) + c_2 \sin(4t) \right] y(t)=e3t[c1sin⁡(4t)−c2cos⁡(4t)] y(t) = e^{3t} \left[ c_1 \sin(4t) - c_2 \cos(4t) \right] Solved on Solvelet with Basic AI Model
Some of the related questions asked by Eleanor Williams on Solvelet
1. Solve the linear system of differential equations: dy1dt=−2y1+y2 \frac{{dy_1}}{{dt}} = -2y_1 + y_2 , dy2dt=3y1−4y2 \frac{{dy_2}}{{dt}} = 3y_1 - 4y_2 , where y1(0)=1 y_1(0) = 1 and y2(0)=2 y_2(0) = 2 .2. Find the general solution to the linear system of differential equations: dxdt=−2x+3y\frac{dx}{dt} = -2x + 3y, dydt=x−y\frac{dy}{dt} = x - y, where x(0)=2x(0) = 2 and y(0)=1y(0) = 1.
DefinitionIn algebra the literal part of a term means the part with the variable, not coefficient of it. It is just the variables and the powers of them. Example: x2y in 5x2y; x2y is the literal part
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