OBJECTIVES

When you finish this section, you should be able to:

1. Solve a separable first-order differential equation (p. 1062)
2. Identify a homogeneous function of degree \(k\) (p. 1063)
3. Use a change of variables to solve a homogeneous first-order differential equation (p. 1064)
4. Solve applied problems (p. 1066)

spanDEFINITIONspan Separable Equation

A first-order differential equation is said to be separable if it can be written in the form \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] {M(x)\,dx+N(y)\,dy=0} } \]

where \(M\) is a function of \(x\) alone, \(N\) is a function of \(y\) alone, and both \(M\) and \(N\) are continuous on their domains.

Steps for Solving a Separable Differential Equation

Step 1 Express the given equation in the differential form \[ \begin{equation*} M(x)\,dx+N(y)\,dy=0 \end{equation*} \]

Step 2 Integrate to obtain the general solution \[ \begin{equation*} \int M(x)\,dx+\int N(y)\,dy=C \end{equation*} \]

where \(C\) is the constant of integration.

Solving a Separable First-Order Differential Equation

Solve \(\dfrac{dy}{dx}=\dfrac{2e^{x}}{y^{2}}\)

Solution Use the steps from above:

Step 1 Express \(\dfrac{dy}{dx}=\dfrac{2e^{x}}{y^{2}}\) in the differential form: \[ y^{2}dy-2e^{x}dx=0\qquad \hbox{or equivalently, as}\qquad y^{2}dy=2e^{x}dx \]

Step 2 Integrate to obtain the general solution. \[ \begin{eqnarray*} \int y^{2}dy &=&\int 2e^{x}dx \\ \dfrac{y^{3}}{3} &=&2e^{x}+C \end{eqnarray*} \]

where \(C\) is a constant. This solution is expressed implicitly. To obtain the explicit form, solve for \(y.\) \[ \begin{eqnarray*} y^{3} &=&6e^{x}+3C \\ y &=&\sqrt[3]{6e^{x}+3C} \end{eqnarray*} \]

NOW WORK

Problems 5 and 15.

spanDEFINITIONspan Homogeneous Function of Degree \(k\)

A function \(f(x,y)\) is said to be a homogeneous function of degree \(k\) in \(x\) and \(y\) if, and only if, for \(t>0\), \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] {f(tx,ty)=t^{k}f(x,y)} } \]

for some real number \(k\).

Identifying a Homogeneous Function of Degree \(k\)

1. \(f(x,y)=3x^{2}-xy+y^{2}\) is a homogeneous function of degree 2, since \[ \begin{eqnarray*} f(tx,ty)& =&3(tx)^{2}-( tx) ( ty) +(ty)^{2}=t^{2}3x^{2}-t^{2}xy+t^{2}y^{2} \\ & =&t^{2}(3x^{2}-xy+y^{2})=t^{2}f(x,y)\quad \hbox{for all }t>0 \end{eqnarray*} \]
2. \(f(x,y)=\sqrt{x+4y}\) is a homogeneous function of degree \(\dfrac{1}{2}\), since \[ \begin{eqnarray*} f(tx,ty)&=&\sqrt{tx+4( ty) }=\sqrt{t(x+4y)}=\sqrt{t}\sqrt{x+4y}\\ &=&t^{1/2}f(x,y)\quad \hbox{for all }t>0 \end{eqnarray*} \]
3. \(f(x,y)=\dfrac{x}{\sqrt{x^{2}-y^{2}}}\) is a homogeneous function of degree \(0\), since \[ \begin{eqnarray*} f(tx,ty)&=&\frac{tx}{\sqrt{(tx)^{2}-(ty)^{2}}}=\frac{tx}{\sqrt{t^{2}(x^{2}-y^{2})}}=\frac{tx}{t\sqrt{x^{2}-y^{2}}}\\ &=&t^{0}f(x,y)\quad \hbox{for all }t>0 \end{eqnarray*} \]
4. \(f(x,y)=x-y^{2}\) is not a homogeneous function, since \[ f(tx,ty)=tx-(ty)^{2}=t(x-ty^{2})\neq t^{k}(x-y^{2}) \] for all \(t>0\) and some \(k\)

NOW WORK

Problem 21.

spanDEFINITIONspan Homogeneous First-Order Differential Equation

A first-order differential equation of the form \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] {M(x,y)\,dx+N(x,y)\,dy=0} }\]

is said to be homogeneous if \(M\) and \(N\) are homogeneous functions of the same degree.

NOTE

The word “homogeneous” has been used in two different ways: to describe a property of a function and to identify a certain type of first-order differential equation. Context will make it clear which meaning applies.

THEOREM Change of Variables for Homogeneous First-Order Differential Equations

If \(M(x,y)\,dx+N(x,y)\,dy=0\) is a homogeneous first-order differential equation, then it can be transformed into a first-order differential equation whose variables are separable by using the substitution \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] {y=xv( x)} } \]

where \(v=v( x) \) is a differentiable function of \(x\).

NOTE

For simplicity of notation, we usually write \(v\) instead of \(v( x)\).

Proof

In the differential equation \(M( x,y)\,dx+N( x,y)\,dy=0,\) we use the substitution \(y=xv,\) where \(v=v( x) \) is a differentiable function of \(x\). Then \(dy=x\,dv+v\,dx\) and \[ \begin{equation*} M(x,y)\,dx+N(x,y)\,dy=M(x,xv)\,dx+N(x,xv)(x\,dv+v\,dx)=0 \end{equation*} \]

Since \(M\) and \(N\) are each homogeneous functions of degree \(k\), it follows that for some number \(k\) \[ \begin{eqnarray*} x^{k}M(1,v)\,dx+x^{k}N(1,v)(x\,dv+v\,dx)& =&0\qquad \color{#0066A7}{{M ( x,}\, {xv} {) =x^{k}M (1,}\, {v} {){,}}}\\ &&\phantom{0}\qquad \color{#0066A7}{{N (x,}\, {xv} {) =x^{k}N(1,}\, {v} {)}} \\ x^{k}\left[ M( 1,v)\,dx+N( 1,v) ( x\,dv+v\,dx) \right] & =&0 \\ M(1,v)\,dx+N(1,v)x\,dv+N(1,v)v\,dx& =& 0\qquad \color{#0066A7}{{\hbox{Divide out } x^{k}.}} \\ \lbrack M(1,v)+vN(1,v)]\,dx+N(1,v)x\,dv& =&0 \\ \frac{dx}{x}+\frac{N(1,v)}{M(1,v)+vN(1,v)}dv& =&0 \end{eqnarray*} \]

provided neither denominator is \(0\). This first-order differential equation is separable.

Steps for Solving a Homogeneous First-Order Differential Equation \({M(x,y)\,dx\,+\, N(x,y)\,dy=0}\)

Step 1 Confirm that the functions \(M\) and \(N\) are homogeneous of the same degree.

Step 2 Let \(y=xv\). Substitute for \(y\) and \(dy=x\,dv+v\,dx.\)

Step 3 Express the new equation in differential form. The variables will be separable.

Step 4 Integrate to obtain the general solution and replace \(v\) by \(\dfrac{y}{x}.\)

Solving a Homogeneous First-Order Differential Equation

Solve the differential equation \((x^{2}-3y^{2})\,dx+2xy\,dy=0\).

Solution Follow the steps for solving a homogeneous first-order differential equation.

Step 1 Both \(x^{2}-3y^{2}\) and \(2xy\) are homogeneous of degree 2. Do you see why?

Step 2 Let \(y=xv\). Then \(dy=x\,dv+v\,dx\). Substitute these into the differential equation: \[ \begin{eqnarray*} (x^{2}-3y^{2})\,dx+2xy\,dy& =&0 \\ [ x^{2}-3( xv) ^{2}]\,dx+2x (xv) (x\,dv+v\,dx)& =&0\qquad \color{#0066A7}{{dy=x\,} {dv=} {+} {v} {dx}} \\ ( x^{2}-x^{2}v^{2})\,dx+2x^{3}v\,dv& =&0 \\ x^{2}(1-v^{2})\,dx+2x^{3}v\,dv& =&0 \end{eqnarray*} \]

Step 3 Then for \(x\neq 0\) and \(v\neq \pm 1,\) we can separate the variables by dividing by \(x^{3}\) and \(1-v^{2}\). \[ \begin{eqnarray*} x^{2} ( 1-v^{2})\,dx+2x^{3}v\,dv &=&0 \\ \hspace{-2pc}\dfrac{x^{2}( 1-v^{2})\,dx}{x^{3}( 1-v^{2}) }+\dfrac{2x^{3}v\,dv}{x^{3}( 1-v^{2}) } &=&0 \\ \dfrac{dx}{x}+\dfrac{2v\,dv}{1-v^{2}} &=&0 \end{eqnarray*} \]

Step 4 Integrate. \[ \begin{eqnarray*} \hspace{-5pc}\int \frac{dx}{x}-\int \frac{2v\,dv}{v^{2}-1} &=&0\\ \ln \left\vert x\right\vert -\ln \left\vert v^{2}-1\right\vert &=&C_{1} \\ \ln \left\vert \dfrac{x}{v^{2}-1}\right\vert &=&C_{1} \end{eqnarray*} \]

Since \(C_{1}\) is a constant, we can write \(C_{1}=\ln C_{2},\) where \(C_{2}\) is a constant. Then \[ \begin{eqnarray*} \ln \left\vert \dfrac{x}{v^{2}-1}\right\vert &=&\ln C_{2} \\ \left\vert \dfrac{x}{v^{2}-1}\right\vert &=&C_{2} \\ \left\vert \dfrac{x}{\dfrac{y^{2}}{x^{2}}-1}\right\vert &=&C_{2}\qquad \color{#0066A7}{{v}} \color{#0066A7}{\hbox{\(=\dfrac{y}{x}\)}}\\ \left\vert \dfrac{x^{3}}{y^{2}-x^{2}}\right\vert & =&C_{2} \\ x^{3}& =&\pm C_{2}( y^{2}-x^{2}) \end{eqnarray*} \]

Since \(\pm C_{2}\) may be either positive or negative, the general solution can be written as \(x^{3}=C( y^{2}-x^{2}) ,\) where \(C\neq 0\) is a constant.

NOW WORK

Problem 37.

Solving a Homogeneous First-Order Differential Equation

Solve the differential equation \(x\,dy+( 2xe^{y/x}-y)\,dx=0\) if \(y=0 \) when \(x=1\).

Solution Follow the steps for solving a homogeneous first-order differential equation:

Step 1 Both \(x\) and \(2xe^{y/x}-y\) are homogeneous of degree 1.

Step 2 Let \(y=xv\). Then \(dy=x\,dv+v\,dx\). Substitute these into the differential equation. \[ \begin{eqnarray*} x\,dy+( 2xe^{y/x}-y)\,dx &=&0 \\ x( x\,dv+v\,dx) +( 2xe^{v}-xv)\,dx &=&0 \\ x^{2}dv+xv\,dx+2xe^{v}dx-xv\,dx &=&0 \\ x^{2}dv+2xe^{v}dx &=&0 \\ x\,dv+2e^{v}dx &=&0 \end{eqnarray*} \]

Step 3 To separate the variables, we divide by \(xe^{v}.\) Then for \(x\neq 0 \), \[ \begin{equation*} \dfrac{2dx}{x}+\dfrac{dv}{e^{v}}=0 \end{equation*} \]

Step 4 Integrate. \[ \begin{eqnarray*} \int \dfrac{2dx}{x}+\int \dfrac{dv}{e^{v}} &=&0 \\ 2\int \dfrac{dx}{x}+\int e^{-v}dv &=&0 \\ 2\ln \left\vert x\right\vert -e^{-v} &=&C \\ 2\ln \left\vert x\right\vert -e^{-y/x} &=&C\qquad \color{#0066A7}{{v}} \color{#0066A7}{{=\dfrac{y}{x}}} \end{eqnarray*} \]

This is the general solution to the differential equation. To find the particular solution, we substitute \(x=1\) and \(y=0\) to find \(C\). \[ \begin{eqnarray*} 2~\ln 1-e^{0} &=&C \\ C &=&-1 \end{eqnarray*} \]

The particular solution is \[ 2~\ln \left\vert x\right\vert -e^{-y/x}=-1 \]

NOW WORK

Problem 41.

Steps for Finding a Family of Orthogonal Trajectories

Step 1 Find the differential equation for the given family.

Step 2 Replace \(y^{\prime} \) in this differential equation by \(-\dfrac{1}{y^{\prime} };\) the resulting equation is the differential equation for the family of orthogonal trajectories.

Step 3 Find the solution of the differential equation obtained in Step 2.

Finding a Family of Orthogonal Trajectories

Find the family of orthogonal trajectories for the one-parameter family \(y^{2}=Cx^{3}\). See Figure 3 for several graphs in the family.

Figure 3 \(y^{2}=Cx^{3}\)

Solution

Step 1 We differentiate \(y^{2}=Cx^{3}\) with respect to \(x\). \[ 2yy^{\prime} =3Cx^{2} \]

Then we eliminate \(C\) using this equation and the equation \(y^{2}=Cx^{3}\) of the family. Since \(C=\dfrac{y^{2}}{x^{3}},\) we find \[ \begin{eqnarray*} 2yy^{\prime} & =&3\frac{y^{2}}{x^{3}}x^{2} \\ y^{\prime} & =&\frac{3y}{2x} \end{eqnarray*} \]

Step 2 Now we replace \(y^{\prime} \) with \(-\dfrac{1}{y^{\prime} },\) to obtain the differential equation for the family of orthogonal trajectories. \[ \begin{eqnarray*} -\dfrac{1}{y^{\prime} }& =&\frac{3y}{2x} \\ y^{\prime} & =&-\frac{2x}{3y} \end{eqnarray*} \]

Figure 4
Blue: \(y^{2}=Cx^{3};\)
Pink: \(x^{2}+\dfrac{3}{2}y^{2}=k\)

Step 3 The differential equation \(\dfrac{dy}{dx}=-\dfrac{2x}{3y}\) is separable and can be written as \[ \begin{equation*} 2x\,dx+3y\,dy=0 \end{equation*} \]

The general solution is \[ \begin{equation*} x^{2}+\frac{3}{2}y^{2}=K \end{equation*} \]

The orthogonal trajectories for the family \(y^{2}=Cx^{3}\) is a family of ellipses \(x^{2}+\dfrac{3}{2}y^{2}=K\), as shown in Figure 4.

NOW WORK

Problem 43.

NEED TO REVIEW?

Differential equations of the form \(\dfrac{{\it dA}}{dt}=kA\) are discussed in Section 5.5, pp. 382-384.

Analyzing a Company 's Sales Growth

The annual sales of a new company are expected to grow at a rate proportional to the difference between the sales at time \(t\) and an upper limit of $\(5\) million. Suppose the sales are $0 initially and are $\(2\) million after \(4\) years of operation. Determine the annual sales at any time \(t\). How long will it take for the annual sales to reach $\(4\) million?

Solution Since there are no sales initially (\(R=0\)), we use equation (4). Then since the upper limit of sales is $\(5\) million, the sales at time \(t\) are \[ \begin{equation*} y(t)=5(1-e^{-kt})\qquad \color{#0066A7}{{M=5}} \end{equation*} \]

Since sales are $\(2\) million after \(4\) years, the boundary condition is \(y(4)=2.\) We use the boundary condition to find the constant \(k\): \[ \begin{eqnarray*} 2& =&5(1-e^{-4k})\qquad \color{#0066A7}{{y(4)=2}}\\ 5e^{-4k}& =&3 \\[2pt] e^{-4k}& =&0.6 \\[4pt] k& =&\frac{\ln 0.6}{-4} = -0.25\ln 0.6 \end{eqnarray*} \]

So, the function \[ \begin{equation*} y(t)=5\big(1-e^{( 0.25\ln 0.6) t}\big) \end{equation*} \]

models the company's sales in year \(t.\)

To find out how long it will take for sales to reach $\(4\) million, we solve the equation \(y( t) =4.\) Then \[ \begin{eqnarray*} 4& =&5\big( 1-e^{( 0.25\ln 0.6) t}\big) \\ 5e^{( 0.25\ln 0.6) t}& =&1 \\[1pt] (0.25\ln 0.6) t & =&\ln 0.2 \\[3pt] t& =&\frac{\ln 0.2}{0.25\ln 0.6}\approx 12.6 \hbox{years} \end{eqnarray*} \]

It will take approximately \(12.6\) years for annual sales to reach $\(4\) million.

NOW WORK

Problem 53.

NOW WORK

Problem 55.