Case 2 of the general solution Solve the equation y′(t) = ky + b in the case that ky + b \u003c 0 and verify that the general solution is y(t) = Ceᵏᵗ − b/k.

Ch. 9 - Differential Equations

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 9 - Differential Equations

Problema 9.4.36

Capitolo 9, Problema 9.4.36

Case 2 of the general solution Solve the equation y′(t) = ky + b in the case that ky + b < 0 and verify that the general solution is y(t) = Ceᵏᵗ − b/k.

Guida verificata passo dopo passo

Recognize that the given differential equation is a first-order linear ordinary differential equation of the form $y'(t) = ky + b$, where $k$ and $b$ are constants.

Rewrite the equation as $y'(t) - ky = b$ to identify it in the standard linear form $y' + p(t)y = q(t)$, where $p(t) = -k$ and $q(t) = b$.

Find the integrating factor $\mu(t)$, which is given by $\mu(t) = e^{\int -k \, dt} = e^{-kt}$.

Multiply both sides of the differential equation by the integrating factor to get $e^{-kt} y'(t) - k e^{-kt} y = b e^{-kt}$, which simplifies to $\frac{d}{dt} \left( e^{-kt} y \right) = b e^{-kt}$.

Integrate both sides with respect to $t$: $\int \frac{d}{dt} \left( e^{-kt} y \right) dt = \int b e^{-kt} dt$, then solve for $y(t)$ to find the general solution in the form $y(t) = C e^{kt} - \frac{b}{k}$, where $C$ is the constant of integration.

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First-Order Linear Differential Equations

These are differential equations of the form y' = ky + b, where k and b are constants. They describe rates of change proportional to the function itself plus a constant term. Understanding their structure is essential for finding general solutions.

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Classifying Differential Equations

General Solution of Non-Homogeneous Equations

The general solution combines the homogeneous solution (y' = ky) and a particular solution to the non-homogeneous equation (y' = ky + b). This approach ensures all possible solutions are captured, typically expressed as y(t) = Ce^{kt} plus a constant term.

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Solutions to Basic Differential Equations

Verification by Substitution

To verify a proposed solution, substitute it back into the original differential equation. This confirms whether the solution satisfies the equation for all t, ensuring correctness and understanding of the solution's behavior.

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Substitution With an Extra Variable

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