Chapter 2

The explicit solution is \(y=\sqrt{\left(3+\cos ^{2} x\right) /\left(1-x^{2}\right)}\). since \(3+\cos ^{2} x > 0\) for all \(x\) we must have \(1-x^{2}>0\) or \(-1 < x < 1 .\) Thus, the interval of definition is (-1,1)

The left-hand derivative of the function at \(x=1\) is \(1 / e\) and the right- hand derivative at \(x=1\) is \(1-1 / e\). Thus, \(y\) is not differentiable at \(x=1\).

Separating variables we have \(d y /\left(\sqrt{1+y^{2}} \sin ^{2} y\right)=d x\) which is not readily integrated (even by a CAS). We note that \(d y / d x \geq 0\) for all values of \(x\) and \(y\) and that \(d y / d x=0\) when \(y=0\) and \(y=\pi,\) which are equilibrium solutions.

First note that $$d(\sqrt{x^{2}+y^{2}})=\frac{x}{\sqrt{x^{2}+y^{2}}} d x+\frac{y}{\sqrt{x^{2}+y^{2}}} d y$$ Then \(x d x+y d y=\sqrt{x^{2}+y^{2}} d x\) becomes $$\frac{x}{\sqrt{x^{2}+y^{2}}} d x+\frac{y}{\sqrt{x^{2}+y^{2}}} d y=d(\sqrt{x^{2}+y^{2}})=d x$$ The left side is the total differential of \(\sqrt{x^{2}+y^{2}}\) and the right side is the total differential of \(x+c\). Thus \(\sqrt{x^{2}+y^{2}}=x+c\) is a solution of the differential equation.

(a) \((i)\) Using Newton's second law of motion, \(F=m a=m d v / d t,\) the differential equation for the velocity \(v\) is $$m \frac{d v}{d t}=m g \sin \theta \quad \text { or } \quad \frac{d v}{d t}=g \sin \theta$$ where \(m g \sin \theta, 0< \theta< \pi / 2,\) is the component of the weight along the plane in the direction of motion. \((i i)\) The model now becomes $$m \frac{d v}{d t}=m g \sin \theta-\mu m g \cos \theta$$ where \(\mu m g \cos \theta\) is the component of the force of sliding friction (which acts perpendicular to the plane) along the plane. The negative sign indicates that this component of force is a retarding force which acts in the direction opposite to that of motion. \((i i i)\) If air resistance is taken to be proportional to the instantaneous velocity of the body, the model becomes $$m \frac{d v}{d t}=m g \sin \theta-\mu m g \cos \theta-k v$$ where \(k\) is a constant of proportionality. (b) \((i)\) With \(m=3\) slugs, the differential equation is $$3 \frac{d v}{d t}=(96) \cdot \frac{1}{2} \quad \text { or } \quad \frac{d v}{d t}=16$$ Integrating the last equation gives \(v(t)=16 t+c_{1} .\) since \(v(0)=0,\) we have \(c_{1}=0\) and so \(v(t)=16 t\). \((i i)\) With \(m=3\) slugs, the differential equation is $$3 \frac{d v}{d t}=(96) \cdot \frac{1}{2}-\frac{\sqrt{3}}{4} \cdot(96) \cdot \frac{\sqrt{3}}{2} \quad \text { or } \quad \frac{d v}{d t}=4$$ In this case \(v(t)=4 t\). (iii) When the retarding force due to air resistance is taken into account, the differential equation for velocity \(v\) becomes $$3 \frac{d v}{d t}=(96) \cdot \frac{1}{2}-\frac{\sqrt{3}}{4} \cdot(96) \cdot \frac{\sqrt{3}}{2}-\frac{1}{4} v \quad \text { or } \quad 3 \frac{d v}{d t}=12-\frac{1}{4} v$$ The last differential equation is linear and has solution \(v(t)=48+c_{1} e^{-t / 12} .\) since \(v(0)=0,\) we find \(c_{1}=-48,\) so \(v(t)=48-48 e^{-t / 12}\).

Separating variables we have \(d y /(\sqrt{y}+y)=d x /(\sqrt{x}+x) .\) To integrate \(\int d x /(\sqrt{x}+x)\) we substitute \(u^{2}=x\) and get \\[\int \frac{2 u}{u+u^{2}} d u=\int \frac{2}{1+u} d u=2 \ln |1+u|+c=2 \ln (1+\sqrt{x})+c\\]. Integrating the separated differential equation we have\\[2 \ln (1+\sqrt{y})=2 \ln (1+\sqrt{x})+c \text { or } \ln (1+\sqrt{y})=\ln (1+\sqrt{x})+\ln c_{1}.\\] Solving for \(y\) we get \(y=\left[c_{1}(1+\sqrt{x})-1\right]^{2}\).

(a) \((i)\) If \(s(t)\) is distance measured down the plane from the highest point, then \(d s / d t=v .\) Integrating \(d s / d t=16 t\) gives \(s(t)=8 t^{2}+c_{2} .\) Using \(s(0)=0\) then gives \(c_{2}=0 .\) Now the length \(L\) of the plane is \(L=50 / \sin 30^{\circ}=100 \mathrm{ft} .\) The time it takes the box to slide completely down the plane is the solution of \(s(t)=100\) or \(t^{2}=25 / 2,\) so \(t \approx 3.54 \mathrm{s}\). \((i i)\) Integrating \(d s / d t=4 t\) gives \(s(t)=2 t^{2}+c_{2} .\) Using \(s(0)=0\) gives \(c_{2}=0,\) so \(s(t)=2 t^{2}\) and the solution of \(s(t)=100\) is now \(t \approx 7.07 \mathrm{s}\). \((\text {iii})\) Integrating \(d s / d t=48-48 e^{-t / 12}\) and using \(s(0)=0\) to determine the constant of integration, we \(\operatorname{obtain} s(t)=48 t+576 e^{-t / 12}-576 .\) With the aid of a CAS we find that the solution of \(s(t)=100,\) or $$\begin{aligned} &100=48 t+576 e^{-t / 12}-576 &\text { or } \quad 0=48 t+576 e^{-t / 12}-676 \end{aligned}$$ is now \(t \approx 7.84 \mathrm{s}\). (b) The differential equation \(m d v / d t=m g \sin \theta-\mu m g \cos \theta\) can be written $$m \frac{d v}{d t}=m g \cos \theta(\tan \theta-\mu)$$. If \(\tan \theta=\mu, d v / d t=0\) and \(v(0)=0\) implies that \(v(t)=0 .\) If \(\tan \theta< \mu\) and \(v(0)=0,\) then integration implies \(v(t)=g \cos \theta(\tan \theta-\mu) t < 0\) for all time \(t\). (c) since \(\tan 23^{\circ}=0.4245\) and \(\mu=\sqrt{3} / 4=0.4330,\) we see that \(\tan 23^{\circ}<0.4330 .\) The differential equation is \(d v / d t=32 \cos 23^{\circ}\left(\tan 23^{\circ}-\sqrt{3} / 4\right)=-0.251493 .\) Integration and the use of the initial condition gives \(v(t)=-0.251493 t+1 .\) When the box stops, \(v(t)=0\) or \(0=-0.251493 t+1\) or \(t=3.976254 \mathrm{s}\). From \(s(t)=-0.125747 t^{2}+t\) we find \(s(3.976254)=1.988119 \mathrm{ft}\). (d) With \(v_{0} >0, v(t)=-0.251493 t+v_{0}\) and \(s(t)=-0.125747 t^{2}+v_{0} t .\) Because two real positive solutions of the equation \(s(t)=100,\) or \(0=-0.125747 t^{2}+v_{0} t-100,\) would be physically meaningless, we use the quadratic formula and require that \(b^{2}-4 a c=0\) or \(v_{0}^{2}-50.2987=0 .\) From this last equality we find \(v_{0} \approx 7.092164 \mathrm{ft} / \mathrm{s} .\) For the time it takes the box to traverse the entire inclined plane, we must have \(0=-0.125747 t^{2}+7.092164 t-100 .\) Mathematica gives complex roots for the last equation: \(t=\) \(28.2001 \pm 0.0124458 i .\) But, for $$0=-0.125747 t^{2}+7.092164691 t-100$$ the roots are \(t=28.1999 \mathrm{s}\) and \(t=28.2004 \mathrm{s} .\) So if \(v_{0} >7.092164,\) we are guaranteed that the box will slide completely down the plane.

We are looking for a function \(y(x)\) such that \\[y^{2}+\left(\frac{d y}{d x}\right)^{2}=1\\]. Using the positive square root gives \\[\frac{d y}{d x}=\sqrt{1-y^{2}} \Longrightarrow \frac{d y}{\sqrt{1-y^{2}}}=d x \Longrightarrow \sin ^{-1} y=x+c\\]. Note that when \(c=c_{1}=0\) and when \(c=c_{1}=\pi / 2\) we obtain the well known particular solutions \(y=\sin x\) \(y=-\sin x, y=\cos x,\) and \(y=-\cos x .\) Note also that \(y=1\) and \(y=-1\) are singular solutions.

(b) For \(|x|>1\) and \(|y|>1\) the differential equation is \(d y / d x=\sqrt{y^{2}-1} / \sqrt{x^{2}-1} .\) Separating variables and integrating, we obtain \(\frac{d y}{\sqrt{y^{2}-1}}=\frac{d x}{\sqrt{x^{2}-1}} \quad\) and \(\quad \cosh ^{-1} y=\cosh ^{-1} x+c\). Setting \(x=2\) and \(y=2\) we find \(c=\cosh ^{-1} 2-\cosh ^{-1} 2=0\) and \(\cosh ^{-1} y=\cosh ^{-1} x .\) An explicit solution is \(y=x\).

Writing the differential equation as \(\frac{d E}{d t}+\frac{1}{R C} E=0\) we see that an integrating factor is \(e^{t / R C}\). Then $$\begin{aligned} \frac{d}{d t}\left[e^{t / R C} E\right] &=0 \\ e^{t / R C} E &=c \\ E &=c e^{-t / R C} \end{aligned}$$ From \(E(4)=c e^{-4 / R C}=E_{0}\) we find \(c=E_{0} e^{4 / R C} .\) Thus, the solution of the initial-value problem is $$E=E_{0} e^{4 / R C} e^{-t / R C}=E_{0} e^{-(t-4) / R C}$$

(a) An integrating factor for \(y^{\prime}-2 x y=-1\) is \(e^{-x^{2}} .\) Thus \\[ \begin{aligned} \frac{d}{d x}\left[e^{-x^{2}} y\right] &=-e^{-x^{2}} \\ e^{-x^{2}} y &=-\int_{0}^{x} e^{-t^{2}} d t=-\frac{\sqrt{\pi}}{2} \operatorname{erf}(x)+c \end{aligned} \\] From \(y(0)=\sqrt{\pi} / 2,\) and noting that erf \((0)=0,\) we get \(c=\sqrt{\pi} / 2 .\) Thus \\[ y=e^{x^{2}}\left(-\frac{\sqrt{\pi}}{2} \operatorname{erf}(x)+\frac{\sqrt{\pi}}{2}\right)=\frac{\sqrt{\pi}}{2} e^{x^{2}}(1-\operatorname{erf}(x))=\frac{\sqrt{\pi}}{2} e^{x^{2}} \operatorname{erfc}(x) \\] (b) Using a CAS we find \(y(2) \approx 0.226339\)

(a) An implicit solution of the differential equation \((2 y+2) d y-\left(4 x^{3}+6 x\right) d x=0\) is \\[y^{2}+2 y-x^{4}-3 x^{2}+c=0\\]. The condition \(y(0)=-3\) implies that \(c=-3 .\) Therefore \(y^{2}+2 y-x^{4}-3 x^{2}-3=0\). (b) Using the quadratic formula we can solve for \(y\) in terms of \(x\): \\[y=\frac{-2 \pm \sqrt{4+4\left(x^{4}+3 x^{2}+3\right)}}{2}\\]. The explicit solution that satisfies the initial condition is then \\[y=-1-\sqrt{x^{4}+3 x^{3}+4}\\]. (c) From the graph of the function \(f(x)=x^{4}+3 x^{3}+4\) below we see that \(f(x) \leq 0\) on the approximate interval \(-2.8 \leq x \leq-1.3 .\) Thus the approximate domain of the function $$y=-1-\sqrt{x^{4}+3 x^{3}+4}=-1-\sqrt{f(x)}$$ is \(x \leq-2.8\) or \(x \geq-1.3 .\) The graph of this function is shown below. (d) Using the root finding capabilities of a CAS, the zeros of \(f\) are found to be -2.82202 and \(-1.3409 .\) The domain of definition of the solution \(y(x)\) is then \(x>-1.3409 .\) The equality has been removed since the derivative \(d y / d x\) does not exist at the points where \(f(x)=0 .\) The graph of the solution \(y=\phi(x)\) is given on the right.

(a) Separating variables and integrating, we have \\[\left(-2 y+y^{2}\right) d y=\left(x-x^{2}\right) d x\\] and \\[-y^{2}+\frac{1}{3} y^{3}=\frac{1}{2} x^{2}-\frac{1}{3} x^{3}+c\\]. Using a CAS we show some contours of \\[f(x, y)=2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}\\]. The plots shown on \([-7,7] \times[-5,5]\) correspond to \(c\) -values of -450,-300,-200,-120,-60,-20,-10,-8.1,-5 \(-0.8,20,60,\) and 120. (b) The value of \(c\) corresponding to \(y(0)=\frac{3}{2}\) is \(f\left(0, \frac{3}{2}\right)=-\frac{27}{4}\) The portion of the graph between the dots corresponds to the solution curve satisfying the intial condition. To determine the interval of definition we find \(d y / d x\) for \\[2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-\frac{27}{4}\\]. Using implicit differentiation we get \(y^{\prime}=\left(x-x^{2}\right) /\left(y^{2}-2 y\right)\) which is infinite when \(y=0\) and \(y=2 .\) Letting \(y=0\) in \(2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-\frac{27}{4}\) and using a CAS to solve for \(x\) we get \(x=-1.13232 .\) Similarly, letting \(y=2,\) we find \(x=1.71299 .\) The largest interval of definition is approximately (-1.13232,1.71299). (c) The value of \(c\) corresponding to \(y(0)=-2\) is \(f(0,-2)=-40\) The portion of the graph to the right of the dot corresponds to the solution curve satisfying the initial condition. To determine the interval of definition we find \(d y / d x\) for \(2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-40\). Using implicit differentiation we get \(y^{\prime}=\left(x-x^{2}\right) /\left(y^{2}-2 y\right)\) which is infinite when \(y=0\) and \(y=2 .\) Letting \(y=0\) in \(2 y^{3}-6 y^{2}+2 x^{3}-3 x^{2}=-40\) and using a CAS to solve for \(x\) we get \(x=-2.29551 .\) The largest interval of definition is approximately \((-2.29551, \infty)\).