First-Order, First-Degree Equations
Variable separable
Example 3.1 (Verification of a solution) A solution must satisfy the ODE. Verify that \(y=c/x\) (\(c\) an arbitrary constant) is a solution of
\(xy'=-y\).
Solution. \(y'=-c/x^2\), so \(xy'=-c/x=-y\). The ODE is satisfied.
Example 3.2 (Graphical solution) \(y'=\dd y/\dd x=\cos x\).
Solution. Separating variables, \(\int\dd y=\int\cos x\,\dd x\), i.e.
\[
y=\int\cos x\,\dd x=\sin x+c,
\]
with \(c\) arbitrary: a family of curves, shown for \(c=-3,-2,\dots,4\).
figureFamily of solutions \(y=\sin x+c\).
Example 3.3 ((A) Exponential growth) \(y'=\dfrac{\dd y}{\dd t}=0.2y\).
Solution. Separating variables, \(y=c\,\ee^{0.2t}\). The ODE \(y'=ky\) with \(k>0\) models exponential growth,
e.g. of bacterial colonies or animal populations. For small human populations in a large country
(e.g. the early United States) it is known as Malthus's law.
Example 3.4 ((B) Exponential decay) \(y'=\dfrac{\dd y}{\dd t}=-0.2y\).
Solution. Similarly, \(y=c\,\ee^{-0.2t}\), which describes, for example, a radioactive substance.
figureSolutions of \(y'=0.2y\) (exponential growth, left) and of \(y'=-0.2y\)
(exponential decay, right).
Example 3.5 (Initial value problem) Solve the initial value problem \(y'=\dfrac{\dd y}{\dd x}=3y\), \(y(0)=5.7\).
Solution. Separating variables gives the general solution \(y=c\,\ee^{3x}\). The initial condition gives
\(y(0)=c=5.7\), so \(y=5.7\,\ee^{3x}\), a particular solution.
Example 3.6 (Radioactivity: exponential decay) \[\frac{\dd N}{\dd t}=-kN,\]
with \(k>0\), so that the minus sign gives decay.
Solution. Separating variables, \(N(t)=N(0)\,\ee^{-kt}\).
Interpretation. \(N(t)\) is the amount of substance at time \(t\). It starts from \(N(0)\),
decreases because \(k>0\), and tends to zero as \(t\to\infty\).
figureRadioactivity (exponential decay).
Singular solution A solution that cannot be obtained from the general solution for any
value of the constant is a singular solution.
Example 3.7 The ODE \(y'^2-xy'+y=0\) is of this kind. Check the general solution \(y=cx-c^2\) and the singular
solution \(y=x^2/4\).
figureParticular solutions and the singular solution.
Solution. For \(y=cx-c^2\), \(y'=c\) and \(c^2-xc+(cx-c^2)=0\). For \(y=x^2/4\), \(y'=x/2\) and
\(\frac{x^2}4-\frac{x^2}2+\frac{x^2}4=0\). No constant \(c\) turns \(cx-c^2\) into \(x^2/4\); the lines
\(y=cx-c^2\) are the tangents to the parabola \(y=x^2/4\) (its envelope).
Example 3.8 Solve the differential equation \(9yy'+4x=0\).
Solution. Separating variables, \(9y\,\dd y=-4x\,\dd x\); integrating,
\[
\frac92y^2=-2x^2+\tilde c,\qquad\text{i.e.}\qquad\frac{x^2}{9}+\frac{y^2}{4}=c\quad(c=\tilde c/18),
\]
a family of ellipses.
Example 3.9 Solve the differential equation \(y'=1+y^2\).
Solution. Separating variables and integrating,
\[
\frac{\dd y}{1+y^2}=\dd x,\qquad \tan^{-1}y=x+c,\qquad y=\tan(x+c).
\]
Example 3.10 (Exponential growth or decay) Solve \(y'=ky\), which governs exponential growth (\(k>0\)) or decay (\(k<0\)).
Solution. The equation is variable separable:
\[
\frac{\dd y}{y}=k\,\dd x,\qquad \ln\abs y=kx+c,
\]
since \((\ln\abs y)'=y'/y\) for \(y>0\) and for \(y<0\). Taking exponentials,
\[
\abs y=\ee^{kx+c}=\ee^{kx}\ee^c,\qquad\text{thus}\qquad y=c\,\ee^{kx},
\]
where the new \(c\) is \(+\ee^c\) for \(y>0\) and \(-\ee^c\) for \(y<0\); \(c=0\) (giving \(y=0\)) is also
admitted. This is the general solution.
Example 3.11 (Initial value problem) Solve the initial value problem \(y'=-\dfrac yx\), \(y(1)=1\).
Solution. Separating and integrating,
\[
\frac{\dd y}{y}=-\frac{\dd x}{x},\qquad \ln\abs y=-\ln\abs x+\tilde c=\ln\frac1{\abs x}+\tilde c.
\]
Taking exponentials, \(y=c/x\). The initial condition gives \(c=1\), so \(y=1/x\).
Example 3.12 (Initial value problem: bell-shaped curves) Solve the initial value problem \(y'=-2xy\), \(y(0)=1\).
Solution. Separating, integrating and taking exponentials,
\[
\frac{\dd y}{y}=-2x\,\dd x,\qquad \ln\abs y=-x^2+\tilde c,\qquad \abs y=\ee^{-x^2+\tilde c}.
\]
With \(c=\pm\ee^{\tilde c}\) (and \(c=0\) for \(y=0\)) the general solution is \(y=c\,\ee^{-x^2}\);
\(y(0)=1\) gives the Gaussian \(y=\ee^{-x^2}\).
figureThe bell-shaped curve \(y=\ee^{-x^2}\).
Homogeneous and reducible equations: reduction to variable-separable form
Some non-separable ODEs become separable when a new unknown function replaces \(y\). Consider
\[
y'=f\Bigl(\frac yx\Bigr).
\]
Put \(y=ux\), so \(y'=u'x+u\). Then \(u'x=f(u)-u\), i.e.
\[
\frac{\dd u}{f(u)-u}=\frac{\dd x}{x}.
\]
Example 3.13 Solve the ODE \(2xyy'=y^2-x^2\).
Solution. Dividing by \(2xy\),
\[
y'=\frac{y^2}{2xy}-\frac{x^2}{2xy}=\frac12\Bigl(\frac yx-\frac xy\Bigr).
\]
With \(y=ux\), \(y'=u'x+u\):
\[
u'x+u=\frac12\Bigl(u-\frac1u\Bigr),\qquad\text{so}\qquad
u'x=-\frac12\Bigl(u+\frac1u\Bigr)=-\frac{u^2+1}{2u}.
\]
Separating variables,
\[
\frac{2u\,\dd u}{1+u^2}=-\frac{\dd x}{x}.
\]
Integrating,
\[
\ln(1+u^2)=-\ln\abs x+c^*=\ln\frac1{\abs x}+c^*,\qquad 1+u^2=\frac cx.
\]
With \(u=y/x\) and multiplying by \(x^2\),
\[
x^2+y^2=cx,\qquad\text{i.e.}\qquad\Bigl(x-\frac c2\Bigr)^2+y^2=\frac{c^2}4.
\]
These are circles with centres on the \(x\) axis, all passing through the origin.
