Chapter 8 Solutions

(a)

1. The real logarithm requires $y>0$. Differentiate both sides of $(x+y{)}^2=\ln y$ with respect to x, treating y as a function of x.
2. The chain rule gives $2(x+y)(1+y\prime )=y\prime /y$.
3. Multiply by y and collect y′ terms: $2y(x+y)=y\prime (1-2y(x+y\left)\right)$.
4. Thus $y\prime =2y(x+y)/(1-2y(x+y\left)\right)$ wherever the denominator is nonzero. At a zero denominator the relation does not define a finite derivative by this formula.

(b)

1. Differentiate $\sin x+{\sin }^{2}y-y=0$. The chain rule gives $\cos x+2\sin y\cos y·y\prime -y\prime =0$.
2. Collect terms and use $2\sin y\cos y=\sin \left(2y\right)$.
3. The result is $y\prime =\cos x/(1-\sin (2y\left)\right)$ where the denominator is nonzero.

(c)

1. Differentiate $e^y+x=0$ to obtain $e^{y}y\prime +1=0$.
2. Therefore $y\prime =-e^{-y}$. Since $e^y=-x>0$, the real curve has x<0 and this is also $y\prime =1/x$.

(a)

1. The highest derivative in $\mathrm{xy}″+y\prime +y^2=0$ is y″, so the equation is second order.
2. The y² term is nonlinear in the dependent variable, so the equation is nonlinear.

(b)

1. The highest derivative in $y\prime -y=x^2$ is y′, so the equation is first order.
2. The dependent variable and its derivative occur only to the first power, with coefficients depending on x. This is a linear equation.

(c)

1. The highest derivative in $y^{\left(5\right)}-\mathrm{yy}\prime +(x+1)y=0$ is the fifth derivative, so the equation is fifth order.
2. The product yy′ is nonlinear in the dependent variable and its derivative. Therefore the equation is nonlinear.

(a)

1. For $y=A\sin x$, differentiation gives $y\prime =A\cos x$ and $y″=-A\sin x$.
2. Since $-A\sin x=-y$, the equation $y″=-y$ holds for every real x and every constant A.

(b)

1. For $y=B\cos x$, differentiation gives $y\prime =-B\sin x$ and $y″=-B\cos x$.
2. Thus $y″=-y$ holds for every real x and every constant B.

(c)

1. For $y=e^x$, both the first and second derivatives are $e^x$.
2. The requested equation would require $e^x=-e^x$, impossible for real x because $e^x>0$. Therefore this function is not a solution.

(d)

1. For $y=x^2$, $y\prime =2x$ and $y″=2$.
2. Substitute: $x^{2}y″-\mathrm{xy}\prime =2x^2-2x^2=0$. Thus the function is a solution for every real x.

(a)

1. The equation $\mathrm{dy}/\mathrm{dx}=\ln x$ is real only for $x>0$. Write $\mathrm{dy}=\ln x\mathrm{dx}$.
2. Integrate by parts on the right: $\int \ln x\mathrm{dx}=x\ln x-x$.
3. Thus $y=x\ln x-x+C$ on x>0. Differentiation returns ln x.

(b)

1. Work on an interval where $\sin x\ne 0$, since the original right-hand side is $y/\sin x$.
2. For y≠0, separate $\mathrm{dy}/y=\mathrm{dx}/\sin x$ and integrate: $\ln \left|y\right|=\ln \left|\tan \right(x/2\left)\right|+C$.
3. Exponentiating and absorbing the sign into a real constant gives $y=C\tan (x/2)$. The constant C=0 also gives the equilibrium solution. The formula is used separately on each interval between consecutive multiples of π.

(c)

1. The original equation contains $1/y$, so y=0 is excluded. Separate $\mathrm{dy}/y=({\cos }^{2}x-x)\mathrm{dx}$.
2. Use ${\cos }^{2}x=(1+\cos 2x)/2$ to integrate: $\ln \left|y\right|=x/2+\sin \left(2x\right)/4-x^2/2+C$.
3. Therefore $y=C\exp (x/2+\sin (2x)/4-x^2/2)$, with $C\ne 0$ because the printed equation excludes y=0.

(a)

1. Treat y and z as constants. Let $N=\sin x\cos \left(\mathrm{xy}\right)$ and $D=\ln \left(\mathrm{xz}\right)$. The real domain requires $\mathrm{xz}>0$ and $\mathrm{xz}\ne 1$.
2. The product and chain rules give $N_x=\cos x\cos \left(\mathrm{xy}\right)-y\sin x\sin \left(\mathrm{xy}\right)$; the denominator derivative is $D_x=1/x$.
3. The quotient rule gives $\left(\right(\cos x\cos \left(\mathrm{xy}\right)-y\sin x\sin \left(\mathrm{xy}\right)\left)\ln \right(\mathrm{xz})-\sin x\cos (\mathrm{xy})/x)/\left(\ln \right(\mathrm{xz}){)}^2$.

(b)

1. The logarithmic term $\ln (z/(2z+y\left)\right)$ has no x dependence. Therefore any mixed derivative containing an x derivative annihilates it, on its real domain $z/(2z+y)>0$.
2. Differentiate $x^3{y}^{2}z$ first in z to get $x^3{y}^2$, then in y to get $2x^{3}y$, then in x to get $6x^{2}y$.
3. Thus the requested mixed derivative is $6x^{2}y$. The functions are smooth on their domain, so other derivative orders give the same result.

(c)

1. The real domain requires $\mathrm{xz}>0$ and $\mathrm{yz}>0$. Treat x and y as constants and use the product rule.
2. Because $\partial \ln \left(\mathrm{xz}\right)/\partial z=1/z$ and $\partial \ln \left(\mathrm{yz}\right)/\partial z=1/z$, the answer is $\left(\ln \right(\mathrm{yz})+\ln (\mathrm{xz}\left)\right)/z$.

(d)

1. Set $u=z^2+y$, requiring $u>0$. The first z derivative is $4z^3\ln u+2z^5/u+x\sin \left(\mathrm{zx}\right)$.
2. Differentiate again: the first term gives $12z^2\ln u+8z^4/u$, the second gives $10z^4/u-4z^6/u^2$, and the last gives $x^2\cos \left(\mathrm{zx}\right)$.
3. Collect the terms to obtain $12z^2\ln (z^2+y)+18z^4/(z^2+y)-4z^6/(z^2+y{)}^2+x^2\cos (\mathrm{zx})$.

1. Differentiate $x^2+y^2=25$ with respect to x: $2x+2yy\prime =0$.
2. Solve for the derivative to obtain $y\prime =-x/y$, valid where y≠0. At y=0 the circle has a vertical tangent.

1. Differentiate $\mathrm{xy}=1$ using the product rule: $y+\mathrm{xy}\prime =0$.
2. Since the original relation excludes x=0, divide by x to get $y\prime =-y/x=-1/x^2$.

1. Differentiate $x^3+y^3=6\mathrm{xy}$: $3x^2+3y^{2}y\prime =6y+6\mathrm{xy}\prime$.
2. Collect y′ terms: $(3y^2-6x)y\prime =6y-3x^2$.
3. Divide where $y^2-2x\ne 0$ to obtain $y\prime =(2y-x^2)/(y^2-2x)$.

1. The real domain of $\ln \left(\mathrm{xy}\right)=x$ requires $\mathrm{xy}>0$, hence x and y are nonzero.
2. Differentiate to get $(y+\mathrm{xy}\prime )/\left(\mathrm{xy}\right)=1$.
3. Multiplying by xy gives $y+\mathrm{xy}\prime =\mathrm{xy}$, hence $y\prime =y(x-1)/x$.

1. Differentiate $\sin y=x^2$: $\cos y·y\prime =2x$.
2. Divide by cos y where it is nonzero. The real relation itself requires $\left|x\right|\le 1$.

1. The highest derivative in $y″+\mathrm{xy}\prime +y=\sin x$ is y″, so the order is 2.
2. Each dependent-variable term is linear in y or a derivative, and the coefficients depend only on x.

1. The highest derivative is y′, so the order is 1.
2. The square $(y\prime {)}^2$ violates linearity, even though the derivative order remains one.

1. The highest derivative is $y‴$, so the order is 3.
2. The dependent variable appears as y², which makes the equation nonlinear.

1. For $y={\mathrm{Ce}}^{-2x}$, the derivative is $y\prime =-2{\mathrm{Ce}}^{-2x}$.
2. Substitute into $y\prime +2y$: $-2{\mathrm{Ce}}^{-2x}+2{\mathrm{Ce}}^{-2x}=0$. This holds for every constant C and every real x.

1. Integrate $y″=6x$ once to obtain $y\prime =3x^2+C_1$.
2. Integrate again to obtain $y=x^3+C_{1}x+C_2$. Two independent constants are needed for the second-order equation.
3. Differentiate twice to check that every member of this family has second derivative 6x.

1. For y≠0, separate $\mathrm{dy}/y=3\mathrm{dx}$.
2. Integration gives $\ln \left|y\right|=3x+C$, so $y={\mathrm{Ce}}^{3x}$.
3. The equilibrium solution y=0 is also included by C=0; these solutions exist for all real x.

1. The original $y\prime =x/y$ requires y≠0. Multiply by y and separate: $y\mathrm{dy}=x\mathrm{dx}$.
2. Integrating gives $y^2/2=x^2/2+C$, or $y^2=x^2+C$.
3. Thus $y=\pm \mathrm{sqrt}(x^2+C)$, with a fixed sign on any interval where $x^2+C>0$. The zero endpoints are excluded from the original equation.

1. For y≠0, separate $\mathrm{dy}/y=2x\mathrm{dx}/(1+x^2)$.
2. The substitution $u=1+x^2$ gives $\ln \left|y\right|=\ln (1+x^2)+C$.
3. Hence $y=C(1+x^2)$. Including C=0 restores the equilibrium solution, and 1+x² is positive for all real x.

1. The equilibrium solutions are y=0 and y=1. For other solutions, separate $\mathrm{dy}/\left(y\right(1-y\left)\right)=\mathrm{dx}$.
2. Use $1/\left(y\right(1-y\left)\right)=1/y+1/(1-y)$. Integration gives $\ln |y/(1-y\left)\right|=x+C$.
3. Solving for y yields $y=1/(1+{\mathrm{Ce}}^{-x})$, on any interval where the denominator is nonzero. This family includes y=1 when C=0; add y=0 separately.

1. The equilibrium y=0 solves the equation. For y≠0, separate $y^{-2}\mathrm{dy}=\mathrm{dx}$.
2. Integration gives $-1/y=x+C$, hence $y=-1/(x+C)$.
3. This nonzero family is valid on intervals excluding $x=-C$. Restore the equilibrium y=0 separately.

1. Separate $\mathrm{dy}/y=x\mathrm{dx}$ for nonzero y and integrate: $\ln \left|y\right|=x^2/2+C$.
2. Thus $y={\mathrm{Ce}}^{x^2/2}$. The initial condition $y\left(0\right)=2$ gives C=2.
3. The resulting positive function is defined for all real x, and its derivative is xy.

1. Divide by the always-positive $1+y^2$ to separate $\mathrm{dy}/(1+y^2)=\cos x\mathrm{dx}$.
2. Integrate to obtain $\arctan y=\sin x+C$.
3. Locally, $y=\tan (\sin x+C)$, on intervals where cosine of the angle is nonzero. The constant can be chosen so the integrated arctangent stays on its principal branch; equivalent constants differ by multiples of π.

1. Multiply through by $1+y^2$ to separate $(1+y^2)\mathrm{dy}=x^3\mathrm{dx}$.
2. Integrate to get $y+y^3/3=x^4/4+C$.
3. The left-hand side has derivative $1+y^2>0$ with respect to y and ranges over all real values. It therefore defines a unique real y for each x and each C, with no division-by-zero restriction.

1. For $f=x^{2}y+3y^2$, treat y as constant when differentiating in x. This gives $f_x=2\mathrm{xy}$.
2. Treat x as constant when differentiating in y. This gives $f_y=x^2+6y$.

1. When differentiating $\sin \left(\mathrm{xy}\right)$ in x, treat y as constant. The inner derivative is y, giving $f_x=y\cos \left(\mathrm{xy}\right)$.
2. In the y derivative, x is constant. The inner derivative is x, giving $f_y=x\cos \left(\mathrm{xy}\right)$.

1. Differentiate $f=x^2{y}^3$ in x to get $f_x=2{\mathrm{xy}}^3$, then in y to get $f_{\mathrm{xy}}=6{\mathrm{xy}}^2$.
2. Reverse the order: $f_y=3x^2{y}^2$, then $f_{\mathrm{yx}}=6{\mathrm{xy}}^2$.
3. The equality is consistent with the polynomial’s continuous second derivatives.

1. For $f=x/(x+y)$, the domain excludes $x+y=0$. In the x derivative, the quotient rule gives $\left(\right(x+y)-x)/(x+y{)}^2=y/(x+y{)}^2$.
2. In the y derivative, the numerator is constant, so $f_y=-x/(x+y{)}^2$.

1. For $f=e^{\mathrm{xy}}$, differentiation in x gives $f_x={\mathrm{ye}}^{\mathrm{xy}}$.
2. Differentiate this in x again, keeping y constant: $f_{\mathrm{xx}}=y^2{e}^{\mathrm{xy}}$.
3. For the mixed derivative, differentiate ${\mathrm{ye}}^{\mathrm{xy}}$ in y using the product rule: $f_{\mathrm{xy}}=e^{\mathrm{xy}}+{\mathrm{xye}}^{\mathrm{xy}}=(1+\mathrm{xy})e^{\mathrm{xy}}$.

1. The real domain of $\ln (x^2+y^2)$ excludes $(x,y)=(0,0)$.
2. Using the chain rule gives $f_x=2x/(x^2+y^2)$ and $f_y=2y/(x^2+y^2)$.

1. Differentiate $f=x^2{\mathrm{yz}}^3$ in z to get $f_z=3x^2{\mathrm{yz}}^2$.
2. Differentiate this in y to get $f_{\mathrm{zy}}=3x^2{z}^2$.
3. Finally differentiate in x to obtain $f_{\mathrm{zyx}}=6{\mathrm{xz}}^2$. This polynomial is smooth, so any ordering of one derivative in each variable gives the same result.