Math Problem Statement
Solution
Let's solve the differential equations one by one as given in the image.
Problem (1)
\frac{dx}{dt} - y = 1, & x|_{t=0} = 1; \\ \frac{dy}{dt} - x = 0, & y|_{t=0} = 1. \end{cases}$$ ### Solution 1. **First Equation:** $$\frac{dx}{dt} = y + 1$$ 2. **Second Equation:** $$\frac{dy}{dt} = x$$ Take the derivative of the second equation with respect to $$t$$: $$\frac{d^2y}{dt^2} = \frac{dx}{dt}$$ From the first equation, substitute $$\frac{dx}{dt} = y + 1$$: $$\frac{d^2y}{dt^2} = y + 1$$ This is a second-order linear differential equation: $$\frac{d^2y}{dt^2} - y = 1$$ ### Homogeneous Equation $$\frac{d^2y}{dt^2} - y = 0$$ The characteristic equation is: $$r^2 - 1 = 0 \implies r = \pm 1$$ Thus, the general solution to the homogeneous equation is: $$y_h(t) = C_1e^t + C_2e^{-t}$$ ### Particular Solution Assume a particular solution of the form $$y_p(t) = A$$: $$0 - A = 1 \implies A = -1$$ Thus, the general solution is: $$y(t) = C_1e^t + C_2e^{-t} - 1$$ ### Initial Conditions From $$y(0) = 1$$: $$1 = C_1 + C_2 - 1 \implies C_1 + C_2 = 2$$ Substitute $$y$$ back into $$\frac{dy}{dt} = x$$ to find $$x(t)$$: $$\frac{dy}{dt} = C_1e^t - C_2e^{-t}$$ $$x(t) = C_1e^t - C_2e^{-t}$$ From $$x(0) = 1$$: $$1 = C_1 - C_2$$ Solve the system: $$C_1 + C_2 = 2 \\ C_1 - C_2 = 1$$ Add these equations: $$2C_1 = 3 \implies C_1 = \frac{3}{2}$$ Subtract these equations: $$2C_2 = 1 \implies C_2 = \frac{1}{2}$$ Thus: $$y(t) = \frac{3}{2}e^t + \frac{1}{2}e^{-t} - 1 \\ x(t) = \frac{3}{2}e^t - \frac{1}{2}e^{-t}$$ Would you like details for any other specific problem or have any questions about this solution? Here are some questions you might want to ask next: 1. Can you solve problem (2) from the image? 2. How do you solve the higher-order differential equations shown? 3. Can you explain the method of undetermined coefficients? 4. What are the solutions for problems (3) and (4)? 5. Can you show the steps to find particular solutions for problem (5)? **Tip:** When solving coupled differential equations, converting them into a single higher-order differential equation can simplify the problem.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Differential Equations
Second-order Differential Equations
Initial Value Problems
Formulas
Second-order linear differential equation
Characteristic equation
Theorems
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Suitable Grade Level
Undergraduate
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