Solve the initial value problem: The differential equation is homogeneous. , . What is the explicit solution ?
Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
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- Combining Functions27m
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- Introduction to Trigonometric Functions38m
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- 1. Limits and Continuity2h 2m
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- 7. Antiderivatives & Indefinite Integrals1h 26m
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- 10. Physics Applications of Integrals 3h 16m
- 11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
- 12. Techniques of Integration7h 39m
- 13. Intro to Differential Equations2h 55m
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7. Antiderivatives & Indefinite Integrals
Initial Value Problems
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Solve the initial value problem: , . What is ?
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Step 1: Recognize that the given problem is an initial value problem involving a first-order differential equation. The equation is x' = 2x, and the initial condition is x(0) = 4.
Step 2: Recall that the general solution to a first-order linear differential equation of the form x' = kx is x(t) = Ce^{kt}, where C is a constant determined by the initial condition.
Step 3: Substitute k = 2 into the general solution formula to get x(t) = Ce^{2t}. This represents the general solution to the differential equation x' = 2x.
Step 4: Use the initial condition x(0) = 4 to solve for the constant C. Substitute t = 0 into x(t) = Ce^{2t}, which gives x(0) = Ce^{0} = C. Since x(0) = 4, we find that C = 4.
Step 5: Substitute C = 4 back into the general solution x(t) = Ce^{2t} to obtain the specific solution x(t) = 4e^{2t}.
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