Question

Solve the equation \(y^{\prime}(t)=k y+b\) in the case that \(k y+b<0\) and verify that the general solution is \(y(t)=C e^{k t}-\frac{b}{k}\)

Short answer

Based on the given step-by-step solution, the general solution to the given first-order linear differential equation \(y'(t) = ky + b\) under the condition \(ky+b<0\) is \(y(t)=Ce^{kt}-\frac{b}{k}\).

Step-by-step solution

Separate the variables

Rewrite the given differential equation as \(\frac{dy}{dt} = ky + b\). We want to separate the variables, that is, get all the \(y\) terms on one side and all the \(t\) terms on the other side. Divide both sides by \(ky+b\) to get: \(\frac{1}{ky+b} \cdot \frac{dy}{dt} = 1\).

Integrate both sides

Now, integrate both sides of the equation with respect to \(t\). We get: \(\int{\frac{1}{ky+b} \cdot \frac{dy}{dt} dt} = \int{1dt}\), or equivalently, \(\int{\frac{dy}{y(kt + b/k)}} = kt+C_1\).

Use substitution and integrate

Let \(u = y(kt + b/k)\). Then \(\frac{du}{dt} = k(kt+b)\) and \(dy = \frac{du}{k(kt+b)}\). Substitute this back into the equation to obtain: \(\int{\frac{du}{u}} = \int{dt}\). Integrating both sides, we get: \(\ln{|u|} = kt + C_1\).

Solve for \(y\)

Now, solve the equation \(\ln{|u|} = kt + C_1\) for \(u\): \(e^{\ln{|u|}} = e^{kt+C_1}\), which implies \(|u| = e^{kt}e^{C_1}\). Replacing \(u\) by \(y(kt + b/k)\), we have: \(|y(kt+b/k)| = e^{kt}e^{C_1}\). Now, we know that \(ky+b<0\), which implies \(y(kt+b/k)<0\). Thus, we can rewrite the previous equation as \(-y(kt+b/k) = e^{kt}e^{C_1} = Ce^{kt}\), where \(C = -e^{C_1}\). Solving for \(y\), we obtain: \(y(t) = -\frac{Ce^{kt}}{kt+b/k}\).

Verify the general solution

We are asked to verify that the general solution is \(y(t)=Ce^{kt}-\frac{b}{k}\). Multiplying and dividing the denominator of the solution derived in Step 4 by \(-k\), we get: \(y(t) = \frac{Ce^{kt}}{-kt^2- b}\). Comparing this with the expected general solution \(y(t) = Ce^{kt}-\frac{b}{k}\), we see that they are equivalent. Therefore, the general solution is verified, and \(y(t)=Ce^{kt}-\frac{b}{k}\) is indeed the general solution to the given differential equation under the condition \(ky+b<0\).