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Chapter 8: Problem 3

Is the equation \(y^{\prime}(t)=2 y-t\) separable?

Short Answer

Expert verified
No, the given differential equation \(y^{\prime}(t)=2 y-t\) is not separable since it cannot be expressed as a product of a function of the independent variable \(t\) and a function of the dependent variable \(y(t)\).

Step by step solution

01

Identify the given differential equation

The given differential equation is \(y^{\prime}(t)=2 y-t\).
02

Rewrite the equation in terms of \(y^{\prime}\)

The equation can be rewritten as \(y^{\prime}(t)-2y(t)=-t\). Let's now analyze if we can separate the equation into a product of two functions.
03

Attempt to separate the variables

At this point, we see that we cannot write the equation as a simple product of a function of \(t\) and a function of \(y(t)\). However, let's try to manipulate the equation in a way to find out if it is separable or not. Rewrite the equation as: \(y^{\prime}(t)=2y(t)-t\) Now we need to check if we can rewrite this as a product of two functions: one in terms of \(t\) and the other in terms of \(y\). We will analyze both sides of the equation. The left side of the equation has one term \(y^{\prime}(t)\), which is the derivative of \(y\) with respect to \(t\). On the right side of the equation, we have two terms: \(2y(t)\) and \(-t\). Unfortunately, there is no obvious way to factor this expression as a product of functions of \(t\) and \(y\) separately. Therefore, we have found that the equation is not separable.
04

Conclusion

The given differential equation, \(y^{\prime}(t)=2 y-t\), is not separable as we cannot express it as a product of a function of the independent variable \(t\) and a function of the dependent variable \(y(t)\).

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Most popular questions from this chapter

a. Show that the logistic growth rate function \(f(P)=r P\left(1-\frac{P}{K}\right)\) has a maximum value of \(\frac{r K}{4}\) at the point \(P=\frac{K}{2}\) b. Show that the Gompertz growth rate function \(f(M)=-r M \ln \left(\frac{M}{K}\right)\) has a maximum value of \(\frac{r K}{e}\) at the point \(M=\frac{K}{e}\)

Consider the general first-order linear equation \(y^{\prime}(t)+a(t) y(t)=f(t) .\) This equation can be solved, in principle, by defining the integrating factor \(p(t)=\exp \left(\int a(t) d t\right) .\) Here is how the integrating factor works. Multiply both sides of the equation by \(p\) (which is always positive) and show that the left side becomes an exact derivative. Therefore, the equation becomes $$p(t)\left(y^{\prime}(t)+a(t) y(t)\right)=\frac{d}{d t}(p(t) y(t))=p(t) f(t).$$ Now integrate both sides of the equation with respect to t to obtain the solution. Use this method to solve the following initial value problems. Begin by computing the required integrating factor. $$y^{\prime}(t)+\frac{1}{t} y(t)=0, \quad y(1)=6$$

Solve the following initial value problems and leave the solution in implicit form. Use graphing software to plot the solution. If the implicit solution describes more than one curve, be sure to indicate which curve corresponds to the solution of the initial value problem. $$y^{\prime}(x)=\sqrt{\frac{x+1}{y+4}}, y(3)=5$$

Consider the following pairs of differential equations that model a predator- prey system with populations \(x\) and \(y .\) In each case, carry out the following steps. a. Identify which equation corresponds to the predator and which corresponds to the prey. b. Find the lines along which \(x^{\prime}(t)=0 .\) Find the lines along which \(y^{\prime}(t)=0\) c. Find the equilibrium points for the system. d. Identify the four regions in the first quadrant of the xy-plane in which \(x^{\prime}\) and \(y^{\prime}\) are positive or negative. e. Sketch a representative solution curve in the xy-plane and indicate the direction in which the solution evolves. $$x^{\prime}(t)=-3 x+6 x y, y^{\prime}(t)=y-4 x y$$

Solve the following initial value problems and leave the solution in implicit form. Use graphing software to plot the solution. If the implicit solution describes more than one curve, be sure to indicate which curve corresponds to the solution of the initial value problem. $$y y^{\prime}(x)=\frac{2 x}{\left(2+y^{2}\right)^{2}}, y(1)=-1$$
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