Kansas State University

Mathematics Department

Textbook Contents

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Change of Variables

Discussion

Basically, there is only one way to solve a first order differential equation. That is to convert it to exact form and integrate it. We have applied this to exact equations, which are already in exact form; to separable equations, which are in exact form after they are separated; and to linear equations, which are in exact form after they are multiplied by an integrating factor. There is one other standard approach to putting a first order equation in exact form, making a change of variables. Two instances where this works are Bernoulli equations and homogeneous equations. (WARNING: The term homogeneous has several different meanings in differential equations. We will encounter the term later with a completely different meaning in Chapter 2.)

Bernoulli Equations

An equation is a Bernoulli equation if it can be written in the form $$ \frac{dy}{dx} + p(x)y = q(x)y^n $$ for some $n$. Bernoulli equations are almost linear equations, they just have an extra $y^n$ term. We can make a change of variables to get rid of this term and rewrite the equation as a linear equation. Let $y = v^{1/(1-n)}$. Then $y^n=v^{n/(1-n)}$ and $$ \frac{dy}{dx} = \frac{1}{1-n}v^{1/(1-n)-1}\frac{dv}{dx} =\frac{1}{1-n}v^{n/(1-n)}\frac{dv}{dx} $$ so plugging into our equation we obtain $$ \frac{1}{1-n}v^{n/(1-n)}\frac{dv}{dx} + p(x)v^{1/(1-n)}=q(x)v^{n/(1-n)} $$ Dividing through by $[1/(1-n)]v^{n/(1-n)}$ we obtain $$ \frac{dv}{dx} + (1-n)p(x)v = (1-n)q(x) $$ which is linear. We now solve the linear equation using ordinary techniques. Finally, since the original problem is stated in terms of $x$ and $y$, the answer should be given in terms of $x$ and $y$ --- not $v$. So we undo our substitution by writing $y=v^{1/(1-n)}$. We must also check for singular solutions of the form $y=0$ since the substitution of $y=v^{1/(1-n)}$ is equivalent to dividing by $y$ if $n>1$.

Paradigm

$$ \frac{dy}{dx}+3y=e^xy^2 $$ STEP 1: Make the substitution $y = v^{1/(1-n)}$.

Here $n=2$ so $y = v^{-1}$ $$ \begin{align} \frac{dy}{dx}&=-v^{-2}\frac{dv}{dx} \\ y^2&=v^{-2} \end{align} $$ and the equation is $$ -v^{-2}\frac{dv}{dx}+3v^{-1}=e^xv^{-2}. $$ STEP 2: Divide through by $v^{-n}$ (or equivalently multiply through by $v^n$ to obtain a linear equation $$ \frac{dv}{dx}-3v=-e^x. $$ STEP 3:Solve the linear equation for v

We follow the paradigm for a linear equation.

  1. $\mu(x) = e^{\int -3\,dx} = e^{-3x}$

  2. $e^{-3x}\frac{dv}{dx}-3e^{-3x}v=-e^{-2x}$

  3. $\frac{d}{dx}(e^{-3x}v)=-e^{-2x}$

  4. $e^{-3x}v=\int -e^{-2x}\,dx=(1/2)e^{-2x}+C$

  5. $v=(1/2)e^x+Ce^{3x}$

STEP 4: Back substitute to find $y$ $$ y=v^{-1}=[(1/2)e^x+Ce^{3x}]^{-1} $$ STEP 5: Check for singular solution $y=0$. $$0+3\cdot0=e^x\cdot0$$ so $y=0$ is a solution as well.

Homogeneous Equations

An equation is homogeneous if it can be written in the form $$ \frac{dy}{dx}=f(y/x) $$ for some function $f$. Usually, it takes some algebraic manipulation to convert the equation to this form. Often, the equation is given in the form $$ \frac{dy}{dx}=\frac{a_nx^n+a_{n-1}x^{n-1}y+a_{n-2}x^{n-2}y^2+\ldots+a_0y^n}{ b_nx^n+b_{n-1}x^{n-1}y+b_{n-2}x^{n-2}y^2+\ldots+b_0y^n} $$ In this case, divide through by $x^n$ to obtain the desired form. Once the equation is in the desired form, we make the change of variables $v=y/x$ so that $y=xv$ and $dy/dx=xdv/dx+v$. Plugging into our equation we obtain $$ x\frac{dv}{dx}+v=f(v) $$ which is separable. We now solve the separable equation using ordinary techniques. Finally, since the original problem is stated in terms of $x$ and $y$, the answer should be given in terms of $x$ and $y$, not $v$. So we undo our substitution by writing $y = xv$.

Paradigm

$$ \frac{dy}{dx}=\frac{2xy}{x^2+y^2} $$ Step 0: Convert to homogeneous form

In this case, every term has order 2 so we divide by $x^2$ to obtain $$ \frac{dy}{dx}=\frac{2(y/x)}{1+(y/x)^2} $$ Step 1: Make the substitution $v=y/x$

This is equivalent to $y=xv$ so we get $dy/dx=x\,dv/dx+v$ and plugging into the equation yields $$ x\frac{dv}{dx}+v=\frac{2v}{1+v^2} $$ Step 2: Solve the separable equation

We use the paradigm for separable equations:

  1. Separate the variables $$ \begin{align} x\frac{dv}{dx}&=\frac{2v}{1+v^2}-v=\frac{v-v^3}{1+v^2} \\ \frac{(1+v^2)dv}{v-v^3}&=\frac{dx}{x} \end{align} $$
  2. Integrate both sides $$ \log(\frac{v}{v^2-1})=\log(x)+C $$
  3. Solve for $v$ $$ \begin{align} \frac{v}{v^2-1}&=kx \\ v&=\frac{1\pm\sqrt{1+4k^2x^2}}{2kx} \end{align} $$
  4. Check for singular solutions.

    We divided by $v^3-v$ which is $0$ at $v=0$ and $v=\pm1$. None of these is included in the general solution so they are all singular solutions.

Step 3: Back Substitute $v=y/x$ to find $y$ $$ y=\frac{1\pm\sqrt{1+4k^2x^2}}{2k} $$ is the general solution and the singular solutions are $$ \begin{align} y&=0 \\ y&=\pm x. \end{align} $$

Example 1: A Bernoulli Initial Value Problem

$$\frac{dy}{dx}+y=\cos(x)/y, \qquad y(0)=1$$ FIRST: Find the general solution.

Step 1: Let $y=v^{1/2}$ then $$(1/2)v^{-1/2}\frac{dv}{dx}+v^{1/2}=\cos(x)/v^{1/2}$$ Step 2: $\displaystyle\frac{dv}{dx}+2v=2\cos(x)$

Step 3: Solving for $v$

  1. $\mu(x)=e^{\int2\,dx}=e^{2x}$

  2. $\displaystyle e^{2x}\frac{dv}{dx}+2e^{2x}v=2e^{2x}\cos(x)$

  3. $\displaystyle \frac{d}{dx}(e^{2x}v)=2e^{2x}\cos(x)$

  4. $\displaystyle e^{2x}v=\int 2e^{2x}\cos(x)\,dx = e^{2x}\left(\frac45\cos(x)+\frac25\sin(x)\right)+C $

  5. $\displaystyle v(x)=\left(\frac45\cos(x)+\frac25\sin(x)\right)+Ce^{-2x}$

Step 4: $\displaystyle y(x)=\sqrt{\frac45\cos(x)+\frac25\sin(x)+Ce^{-2x}} $

Step 5: $y=0$ is not a solution (can't have $\cos(x)/0$ on right-hand-side).

SECOND: Plug in the initial value and solve for the arbitrary constant. $$ \begin{align} y(0)=\sqrt{\frac45\cos(0)+\frac25\sin(0)+Ce^0}&{\buildrel \text{set} \over = } \sqrt{\frac45+C}&=1 \\ C&=\frac15 \end{align} $$ So $\displaystyle y(x)=\sqrt{\frac45\cos(x)+\frac25\sin(x)+\frac15e^{-2x}} $

Example 2: A Homogeneous Initial Value Problem

$$ \frac{dy}{dx}=\frac{3y^2+4xy+2x^2}{2xy+x^2},\qquad y(1)=2 $$ FIRST: Find the general solution:

Step 0: Divide top and bottom by $x^2$ to have everything in terms of $y/x$. $$ \frac{dy}{dx}=\frac{3(y/x)^2+4(y/x)+2}{2(y/x)+1} $$ Step 1: Let $v=y/x$ (or $y=xv$ to get $$ x\frac{dv}{dx}+v=\frac{3v^2+4v+2}{2v+1} $$ Step 2: We now solve this equation using the separable paradigm.

  1. $$\begin{align} x\frac{dv}{dx}&=\frac{3v^4v+2}{2v+1}-v \\ x\frac{dv}{dx}&=\frac{v^2+3v+2}{2v+1} \\ \frac{2v+1\,dv}{v^2+3v+2}&=\frac{dx}{x} \end{align}$$
  2. Partial fractions gives us $\displaystyle \frac{2v+1}{v^2+3v+2}=\frac{3}{v+2}-\frac{1}{v+1}$, so $$\begin{align} \int\frac{2v+1\,dv}{v^2+3v+2}&=\int\frac{dx}{x} \\ 3\int\frac{dv}{v+2}-\int\frac{dv}{v+1}&=\int\frac{dx}{x} \\ 3\log(v+2)-\log(v+1)&=\log(x)+C \\ (v+2)^3&=kx(v+1). \end{align}$$
Step 3: Back-substitute $v=y/x$ and to get $(y/x+2)^3=kx(y/x+1)$. We can multiply through by $x^3$ to simplify this to $$(y+2x)^3=kx^3(y+x).$$ SECOND: Now that we have the general solution, we can use the initial values $y(1)=2$ to find the constant $k$. Plugging in $x=1$ and $y=2$ we get $$\begin{align} (2+2)^3&=k(2+1) \\ 64&=3k \\ 64/3=k \end{align}$$ which gives us our final answer $$ 3(y+2x)^2=64x^3(y+x). $$

Example 3: A Homogeneous Initial Value Problem (Singular Solution)

$$\frac{dy}{dx}=\frac{40x-11y}{2x-y},\qquad y(1)=8$$ FIRST: Find the general solution:

Step 0: Divide top and bottom by $x$ to have everything in terms of $y/x$. $$ \frac{dy}{dx}=\frac{40-11(y/x)}{2-(y/x)} $$ Step 1: Let $v=y/x$ (or $y=xv$) to get $$ x\frac{dv}{dx}+v=\frac{40-11v}{2-v} $$ Step 2: We now solve this equation.

  1. $$ \begin{align} x\frac{dv}{dx}&=\frac{40-11v}{2-v}-v \\ x\frac{dv}{dx}&=\frac{40-13v+v^2}{2-v} \\ \frac{2-v\,dv}{40-13v+v^2}&=\frac{dx}{x} \end{align} $$
  2. Partial fractions tells us that $\displaystyle \frac{2-v}{40-13v+v^2}=\frac{1}{v-5}-\frac{2}{v-8}$, so $$ \begin{align} \int\frac{2-v\,dv}{40-13v+v^2}&=\int\frac{dx}{x} \\ \int\frac{dv}{v-5}-\int\frac{2\,dv}{v-8}&=\int\frac{dx}{x} \\ \log|v-5|-2\log|v-8|&=\log|x| + C \\ v-5 &= kx(v-8)^2. \end{align} $$
Step 3: Plugging in $v=y/x$ we get $(y/x - 5)=kx(y/x - 8)^2$. We can multiply through by $x$ to simplify this to $$y-5x=k(y-8x)^2.$$ SECOND: Now that we have the general solution, we can plug in the initial value to find the constant. Plugging in $x=1$ and $y=8$ we get $3=k\times0$, and that is impossible. This is our cue that we need to go back and find the singular solution.

We divided by $40-13v+v^2=(v-5)(v-8)$ whose roots are $v=5$ and $v=8$. These are both solutions to the equation, though $v-5$ is already included in the general solution with $k=0$. But $v=8$ is a singular solution. Then remembering that $v=y/x$, we find $y/x = 8$ or $$y=8x.$$ This satisfies our initial condition $y(1)=8$, so this is the solution to the initial value problem.

Randomly Generated Examples

A randomly generated Bernoulli problem is below, and a randomly generated homogeneous problem is below that. As always, once you submit an answer, you will have a link to see the detailed solution for the problem, as well as the option to generate a new problem and you can just quickly enter a "dummy" answer like $y=x$ to quickly get to the detailed solution if that is all you want to see.

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