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Chapter 8: Problem 8

Find the general solution of the following equations. $$y^{\prime}(x)=2 y+6$$

Short Answer

Expert verified
Based on the given differential equation \(y^{\prime}(x)=2 y+6\), find the general solution. Solution: The general solution for the given differential equation is \(y(x) = 3 + Ce^{2x}\), where C is an arbitrary constant.

Step by step solution

01

Identify P(x) and Q(x)

In the given differential equation $$y^{\prime}(x)=2 y+6$$, we can rewrite it as $$\frac{dy}{dx} + (-2)y = -6$$. Now, we compare it with the standard form \(\frac{dy}{dx} + P(x)y = Q(x)\) and identify P(x) and Q(x). We get \(P(x) = -2\) and \(Q(x) = -6\).
02

Calculate the integrating factor

The integrating factor is given by $$I(x) = e^{\int P(x) dx}.$$ Plug in P(x) and we get $$I(x) = e^{\int -2 dx} = e^{-2x}.$$
03

Multiply the differential equation by the integrating factor

Now, multiply the given differential equation by the integrating factor (\(e^{-2x}\)): $$e^{-2x} \frac{dy}{dx} + (-2e^{-2x}) y = (-6e^{-2x}).$$
04

Integrate the modified equation

The left side of the modified equation is the derivative of the product \(ye^{-2x}\): $$\frac{d}{dx}(ye^{-2x}) = e^{-2x} \frac{dy}{dx} + (-2e^{-2x}) y.$$ So, we integrate both sides of the equation: $$\int \frac{d}{dx}(ye^{-2x}) dx = \int -6e^{-2x} dx.$$ The left side becomes \(ye^{-2x}\). For the right side, let's use the substitution method: Let \(u = -2x\), then \(\frac{du}{dx} = -2 \Rightarrow dx = -\frac{1}{2} du\). Now we have: $$ye^{-2x} = -\frac{1}{2}\int -6e^{u} du = 3\int e^{u} du.$$ Integrate with respect to u: $$ye^{-2x} = 3e^{u} + C = 3e^{-2x} + C.$$
05

Solve for the general solution

Finally, we solve for y(x) to obtain the general solution: $$y(x) = 3 + Ce^{2x}.$$

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Most popular questions from this chapter

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 \(y(0)=1\). $$y^{\prime}(t)=y(2-y)$$

For the following separable equations, carry out the indicated analysis. a. Find the general solution of the equation. b. Find the value of the arbitrary constant associated with each initial condition. (Each initial condition requires a different constant.) c. Use the graph of the general solution that is provided to sketch the solution curve for each initial condition. $$\begin{array}{l} e^{-y / 2} y^{\prime}(x)=4 x \sin x^{2}-x ; y(0)=0 \\ y(0)=\ln \left(\frac{1}{4}\right), y(\sqrt{\frac{\pi}{2}})=0 \end{array}$$

Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection. A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. Use the following steps to find the orthogonal trajectories of the family of ellipses \(2 x^{2}+y^{2}=a^{2}\) a. Apply implicit differentiation to \(2 x^{2}+y^{2}=a^{2}\) to show that $$ \frac{d y}{d x}=\frac{-2 x}{y} $$ b. The family of trajectories orthogonal to \(2 x^{2}+y^{2}=a^{2}\) satisfies the differential equation \(\frac{d y}{d x}=\frac{y}{2 x} .\) Why? c. Solve the differential equation in part (b) to verify that \(y^{2}=e^{C}|x|\) and then explain why it follows that \(y^{2}=k x\) Therefore, the family of parabolas \(y^{2}=k x\) forms the orthogonal trajectories of the family of ellipses \(2 x^{2}+y^{2}=a^{2}\)

Solve the following initial value problems. When possible, give the solution as an explicit function of \(t\) $$y^{\prime}(t)=\frac{y+3}{5 t+6}, y(2)=0$$

Determine whether the following equations are separable. If so, solve the initial value problem. $$\frac{d y}{d t}=t y+2, y(1)=2$$
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