Question

In Problems 1–22, solve the given differential equation by separation of variables.

$\frac{\text{dN}}{\text{dt}}{\text{+ N = N te}}^{\text{t + 2}}$

Short answer

$\ln \left|N\right|=t{e}^{t+2}-e^{t+2}-t+C$

Step-by-step solution

Definition of separable equation

A first-order differential equation of the form $\frac{dy}{dx}=g\left(x\right)h\left(y\right)$is said to be separable or to have separable variables.

Separate the variables

$\frac{dN}{dt}+N=Nt{e}^{t+2}$

Rearrange

$\frac{dN}{dt}=Nt{e}^{t+2}-N$

Factor the right side

$\frac{dN}{dt}=N\left(t{e}^{t+2}-1\right)$

Separate the variables

$\frac{dN}{N}=\left(t{e}^{t+2}-1\right)dt$

Integration

Integrate both sides

$$\begin{gathered} \int \frac{dN}{N}=\int \left(t{e}^{t+2}-1\right)dt \\ \int \frac{dN}{N}=\int t{e}^{t+2}dt-\int dt······\left(1\right) \end{gathered}$$

Integrating $\int t{e}^{t+2}dt$parts

$u=t\Rightarrow du=dt$

$dv=e^{t+2}\Rightarrow v=e^{t+2}$

So,

$$\begin{gathered} \int t{e}^{t+2}dt=t{e}^{t+2}-\int e^{t+2}dt \\ \int t{e}^{t+2}dt=t{e}^{t+2}-e^{t+2} \end{gathered}$$

Substitute this value in (1) and integrate,

$\ln \left|N\right|=t{e}^{t+2}-e^{t+2}-t+C$

Therefore the solution is${\text{ln |N| = te}}^{\text{t + 2}}{\text{-e}}^{\text{t + 2}}\text{- t + C}$