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Chapter 8: Problem 9

Find the general solution of the following equations. $$u^{\prime}(t)+12 u=15$$

Short Answer

Expert verified
Question: Find the general solution of the first-order linear differential equation $$u'(t) + 12u(t) = 15$$. Answer: The general solution of the given differential equation is $$u(t) = \frac{5}{4} + \frac{C}{e^{12t}}$$.

Step by step solution

01

Identify P(t) and Q(t) in the given linear differential equation

In the given equation, $$u'(t) + 12u(t) = 15$$, we can identify the functions P(t) and Q(t) as follows: - P(t) = 12 (coefficient of the function u(t)) - Q(t) = 15
02

Find the integrating factor (IF)

To find the integrating factor (IF), we need to calculate $$\rho(t)=e^{\int P(t)dt}$$. In our case, we have: $$\rho(t)=e^{\int 12dt} = e^{12t}$$
03

Multiply both sides of the given differential equation by the integrating factor

Now, we will multiply both sides of the equation $$u'(t) + 12u(t) = 15$$ by the integrating factor $$e^{12t}$$. This results in: $$e^{12t}u'(t) + 12e^{12t}u(t) = 15e^{12t}$$
04

Observe that the left-hand side of the equation is the derivative of (IF 脳 u(t))

By noticing that the left-hand side of the equation matches the derivative of '$$u(t)e^{12t}$$', that is: $$\frac{d}{dt}(u(t)e^{12t}) = u'(t)e^{12t} + 12u(t)e^{12t}$$ We can rewrite the equation as: $$\frac{d}{dt}(u(t)e^{12t}) = 15e^{12t}$$
05

Integrate both sides with respect to t

Now, we will integrate both sides of the equation with respect to t: $$\int \frac{d}{dt}(u(t)e^{12t}) dt = \int 15e^{12t} dt$$ $$u(t)e^{12t} = \frac{15}{12}e^{12t} + C$$
06

Solve for u(t) to obtain the general solution

Finally, we will solve for u(t) by dividing both sides by $$e^{12t}$$: $$u(t) = \frac{15}{12} + \frac{C}{e^{12t}}$$ $$u(t) = \frac{5}{4} + \frac{C}{e^{12t}}$$ This is the general solution of the given differential equation.

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Most popular questions from this chapter

a. Show that for general positive values of \(R, V, C_{i},\) and \(m_{0},\) the solution of the initial value problem $$m^{\prime}(t)=-\frac{R}{V} m(t)+C_{i} R, \quad m(0)=m_{0}$$ is \(m(t)=\left(m_{0}-C_{i} V\right) e^{-R t / V}+C_{i} V\) b. Verify that \(m(0)=m_{0}\) c. Evaluate \(\lim m(t)\) and give a physical interpretation of the result. d. Suppose \(^{t} \vec{m}_{0}^{\infty}\) and \(V\) are fixed. Describe the effect of increasing \(R\) on the graph of the solution.

The Gompertz growth equation is often used to model the growth of tumors. Let \(M(t)\) be the mass of a tumor at time \(t \geq 0 .\) The relevant initial value problem is $$ \frac{d M}{d t}=-r M \ln \left(\frac{M}{K}\right), M(0)=M_{0} $$ a. Graph the growth rate function \(R(M)=-r M \ln \left(\frac{M}{K}\right)\) (which equals \(M^{\prime}(t)\) ) assuming \(r=1\) and \(K=4 .\) For what values of \(M\) is the growth rate positive? For what value of \(M\) is the growth rate a maximum? b. Solve the initial value problem and graph the solution for \(r=1, K=4,\) and \(M_{0}=1 .\) Describe the growth pattern of the tumor. Is the growth unbounded? If not, what is the limiting size of the tumor? c. In the general solution, what is the meaning of \(K ?\) where \(r\) and \(K\) are positive constants and \(0 < M_{0} < K\)

Let \(y(t)\) be the concentration of a substance in a chemical reaction (typical units are moles/liter). The change in the concentration, under appropriate conditions, is modeled by the equation \(\frac{d y}{d t}=-k y^{n},\) where \(k>0\) is a rate constant and the positive integer \(n\) is the order of the reaction. a. Show that for a first-order reaction \((n=1)\), the concentration obeys an exponential decay law. b. Solve the initial value problem for a second-order reaction \((n=2)\) assuming \(y(0)=y_{0}\) c. Graph the concentration for a first-order and second-order reaction with \(k=0.1\) and \(y_{0}=1\)

A differential equation of the form \(y^{\prime}(t)=f(y)\) is said to be autonomous (the function \(f\) depends only on \(y\) ). The constant function \(y=y_{0}\) is an equilibrium solution of the equation provided \(f\left(y_{0}\right)=0\) (because then \(y^{\prime}(t)=0\) and the solution remains constant for all \(t\) ). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations. a. Find the equilibrium solutions. b. Sketch the direction field, for \(t \geq 0\). c. Sketch the solution curve that corresponds to the initial condition \(y(0)=1\). $$y^{\prime}(t)=6-2 y$$

Determine whether the following equations are separable. If so, solve the initial value problem. $$\frac{d y}{d x}=e^{x-y}, y(0)=\ln 3$$
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