Question

Solve the first-order linear differential equation of the form dy/dt +ay=g(t), y(0)=y₀ , where y=y(t) and a is a constant.

Answer 1

The solution to the first-order linear differential equation
`\( (dy)/(dt) + ay = g(t) \), with initial condition \( y(0) = y₀ \), is given by \( y(t) = e^(-at) \left( y₀ + \int_(0)^(t) e^(as)g(s) \, ds \right) \).`

Step-by-step explanation:

To solve the given first-order linear differential equation, we can use an integrating factor. The general form of the equation is
`\( (dy)/(dt) + ay = g(t) \)`, where (a) is a constant. Multiply both sides by the integrating factor
`\( e^(at) \)` to obtain:

`\[ e^(at)(dy)/(dt) + ae^(at)y = e^(at)g(t) \]`

This expression can be written as the derivative of
`\( e^(at)y \)` with respect to (t):

`\[ (d)/(dt)(e^(at)y) = e^(at)g(t) \]`

Now, integrate both sides with respect to ( t ) to solve for ( y ):

`\[ e^(at)y = y₀ + \int_(0)^(t) e^(as)g(s) \, ds \]`

Finally, multiply both sides by
`\( e^(-at) \) to isolate \( y \)`:

`\[ y(t) = e^(-at) \left( y₀ + \int_(0)^(t) e^(as)g(s) \, ds \right) \]`

This is the solution to the given differential equation with the initial condition ( y(0) = y₀). The term involving the integral represents the contribution of the forcing function (gt) over the interval [0, t].

Answer 2

The general solution to the first-order linear differential equation dy/dt + ay = g(t) with initial condition y(0) = y₀ is y(t) = e^(-at) * [y₀ + ∫(e^(at) * g(t)) dt].

Step-by-step explanation:

The given first-order linear differential equation dy/dt + ay = g(t) represents a standard form of a linear ordinary differential equation (ODE). To solve this, we use an integrating factor to separate variables and integrate.

First, we identify the integrating factor, which is the exponential function e^(∫a dt). Integrating factor = e^(∫a dt) = e^(at).

To solve the differential equation, we multiply both sides by the integrating factor:

e^(at) * dy/dt + ae^(at) * y = e^(at) * g(t).

This expression can be rewritten using the product rule for differentiation as (e^(at) * y)' = e^(at) * g(t).

Integrating both sides with respect to t gives us:

∫(e^(at) * dy/dt + ae^(at) * y) dt = ∫e^(at) * g(t) dt.

This simplifies to:

e^(at) * y = ∫e^(at) * g(t) dt + C.

Now, solving for y, we get:

y(t) = e^(-at) * [∫e^(at) * g(t) dt + C].

Applying the initial condition y(0) = y₀ allows us to solve for the constant C. Substituting t = 0 and y = y₀ into the equation gives:

y₀ = e^0 * [∫e^(0) * g(t) dt + C],

y₀ = [∫g(t) dt + C].

Therefore, the solution to the first-order linear differential equation with the initial condition y(0) = y₀ is y(t) = e^(-at) * [y₀ + ∫(e^(at) * g(t)) dt].