18. dN/dt+N=Nte^{t+2} by separable method.

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18. dN/dt+N=Nte^{t+2} by separable method.

To solve the differential equation \frac{dN}{dt}+N=Nte^{t+2} by separable method, we need to separate the variables N and t on opposite sides of the equation and integrate both sides.

Starting with the given differential equation:

    \begin{equation*} \frac{dN}{dt} + N = Nte^{t+2} \end{equation*}

We can first solve the homogeneous equation \frac{dN}{dt}+N=0 to get the complementary solution:

    \begin{align*} \frac{dN}{dt}+N &= 0 \end{align*}

    \begin{align*} \frac{dN}{N} &= -dt \end{align*}

    \begin{align*} ln|N| &= -t + C_1 \end{align*}

    \begin{align*}N &= Ce^{-t} \end{align*}

where C is a constant determined by the value of the constant of integration C_1.

Now we need to find a particular solution to the non-homogeneous equation. We can use the method of variation of parameters to do so. Let N=uv, where u and v are functions of t. Then, we have:

    \begin{align*} \frac{dN}{dt} &= \frac{du}{dt}v + u\frac{dv}{dt} = u\frac{dv}{dt} + uv\frac{du}{dt} = u\frac{dv}{dt} + v\left(\frac{du}{dt}+u\right) \end{align*}

Substituting this into the original differential equation and simplifying, we get:

    \begin{align*} u\frac{dv}{dt} &= t e^{t+2} \ \int u , \end{align*}

    \begin{align*} dv &= \int t e^{t+2}dt , \end{align*}

    \begin{align*} uv &= \frac{1}{2}te^{t+2} - \frac{1}{2}e^{t+2} + C_2 \end{align*}

where C_2 is a constant of integration.

Now, we can substitute the complementary and particular solutions back into the original equation to get the general solution:

    \begin{align*} N &= Ce^{-t} + \frac{1}{2}te^{t+2} - \frac{1}{2}e^{t+2} + C_2 \ \end{align*}

where C and C_2 are constants determined by the initial conditions.

Therefore, the general solution to the given differential equation is:

    \begin{equation*} N = Ce^{-t} + \frac{1}{2}te^{t+2} - \frac{1}{2}e^{t+2} + C_2 \end{equation*}

where C and C_2 are constants determined by the initial conditions.

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