 # Section 2.4 – Differences Between Linear and Nonlinear Equations

Directions. The following are review problems for the section. It is recommended you work the problems yourself, and then click "Answer" to check your answer. If you do not understand a problem, you can click "Video" to learn how to solve it.
1. Determine an interval in which the solution of the following initial value problem is certain to exist.
$(t^2-1)y'+(\sin t)y=\frac{\cot t}{t^2-4t+3},\qquad y(2)=-1$

$$I=(1,3)$$

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2. State where in the $$ty$$-plane the hypothesis of the Existence and Uniqueness theorem are satisfied for the following differential equations.
1. $$y'=\dfrac{\ln(ty)}{1-(t^2+y^2)}$$
2. $$y'=(t^2-y)^{1/3}$$

1. Anywhere in the first or third quadrant, but not on the unit circle.
2. Any point off the parabola $$y=t^2.$$

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3. Solve the following initial value problems  and determine  how the interval in which the solution exists depends on the initial value $$y_0$$.
1. $$\displaystyle y'=\frac{-4}{t}y,\qquad$$ $$y(2)=y_0$$
2. $$\displaystyle y'+y^3=0\qquad$$ $$y(t_0)=y_0$$

1. The solution is $$y=\dfrac{16y_0}{t^4}$$ and $$I.V.=(0,\infty)$$. Note that if $$y_0=0$$, then $$y=0$$ is an equilibrium solution.
2. If $$y_0=0$$, then $$y=0$$ is an equilibrium solution with $$I=(-\infty,\infty)$$. If $$y_0>0$$, then $$y=\dfrac{1}{\sqrt{2(t-t_0)+\frac{1}{y_0^2}}}$$ with $$I=\left(t_0-\dfrac{1}{2{y_0}^2},\infty\right).$$ If $$y_0<0$$, then $$y=-\dfrac{1}{\sqrt{2(t-t_0)+\frac{1}{y_0^2}}}$$ with $$I=\left(t_0-\dfrac{1}{2{y_0}^2},\infty\right).$$

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4. Verify that both $$y_1=1-t$$ and $$\displaystyle y_2=\frac{-t^2}{4}$$ are solutions to the same initial value problem $y'(t)=\frac{-t+\sqrt{t^2+4y}}{2},\qquad y(2)=-1.$Does it contradict the existence and uniqueness theorem?

There is no contradiction because the conditions of the existence and uniqueness theorem are only satisfied for points above the parabola $$y=-\dfrac{t^2}{4}$$, and the point $$(2,1)$$ is on the parabola. For verification that $$y_1$$ and $$y_2$$ are solutions to the same initial value problem, see video below.
Video Errata: There is a mistake at the 11:00 mark when the presenter wrote out $$f_y$$, we should have that $$f_y=\dfrac{1}{2}\cdot\dfrac{1}{2}(t^2+4y)^{-\frac{1}{2}}\cdot (4)=\dfrac{1}{\sqrt{t^2+4y}}.$$ This does not change our answer as the condition on $$f_y$$ is still $$t^2+4y>0.$$

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