Exercises — Bernoulli equations — substitution

A Bernoulli equation must have the form $y^{\prime }+P(x)y=Q(x)y^n$ with the nonlinearity living only in a single power $y^n$, and $n\ne 0,1$.

(a) Bernoulli: $P=3$, $Q=5$, $n=4$. ✔ (b) Bernoulli: $P=\frac{2}{x}$, $Q=x^3$, $n=-2$. ✔ (negative $n$ is allowed — $n\in \mathbb{R}$) (c) Not Bernoulli in the interesting sense: $n=0$ here (the right side $x=x\cdot y^0$), which is already linear. No substitution needed. (d) Bernoulli: $P=\sin x$, $Q=1$, $n=\frac{1}{3}$. ✔

(a) $1-n=1-4=-3$, so $v=y^{-3}$. (b) $1-n=1-(-2)=3$, so $v=y^3$. (d) $1-n=1-\frac{1}{3}=\frac{2}{3}$, so $v=y^{2/3}$.

Identify: $P=\frac{1}{x}$, $Q=x$, $n=3$, so $1-n=-2$ and $v=y^{-2}$.

Master result: $v^{\prime }+(1-n)Pv=(1-n)Q$ gives $v^{\prime }+(-2)\frac{1}{x}v=(-2)x⟹v^{\prime }-\frac{2}{x}v=-2x.$ Integrating factor: $\mu =e^{\int -2/xdx}=e^{-2\ln x}=x^{-2}.$ Multiply: $(x^{-2}v{)}^{\prime }=-2x\cdot x^{-2}=-\frac{2}{x}.$ Integrate: $x^{-2}v=-2\ln |x|+C\Rightarrow v=x^2(C-2\ln |x|).$ Back-substitute $v=y^{-2}$: $\boxed{\frac{1}{y^2}=x^2(C-2\ln |x|)}$

Identify: $P=-1$, $Q=x$, $n=2$, $1-n=-1$, $v=y^{-1}$. $v^{\prime }+(-1)(-1)v=(-1)x⟹v^{\prime }+v=-x.$ IF: $\mu =e^{\int 1dx}=e^x.$ Then $(e^{x}v{)}^{\prime }=-x{e}^x.$ Integrate by parts: $\int -x{e}^{x}dx=-(x-1)e^x+C=(1-x)e^x+C.$ So $e^{x}v=(1-x)e^x+C\Rightarrow v=(1-x)+C{e}^{-x}.$ Back-substitute $v=1/y$: $\boxed{\frac{1}{y}=1-x+C{e}^{-x}}$

Identify: $P=1$, $Q=\frac{1}{2}$, $n=3$, $1-n=-2$, $v=y^{-2}$. $v^{\prime }+(-2)(1)v=(-2)\frac{1}{2}⟹v^{\prime }-2v=-1.$ IF: $\mu =e^{-2x}$. Then $(e^{-2x}v{)}^{\prime }=-e^{-2x}.$ Integrate: $e^{-2x}v=\frac{1}{2}{e}^{-2x}+C\Rightarrow v=\frac{1}{2}+C{e}^{2x}.$ Apply IC: $y(0)=1\Rightarrow v(0)=1^{-2}=1$. So $1=\frac{1}{2}+C\Rightarrow C=\frac{1}{2}.$ Thus $v=\frac{1}{2}+\frac{1}{2}{e}^{2x}=\frac{1}{2}(1+e^{2x}).$ Back-substitute $v=y^{-2}$: $\boxed{y^2=\frac{2}{1+e^{2x}}}⟹y=\sqrt{\frac{2}{1+e^{2x}}}$ (positive root since $y(0)=1>0$). Sanity: at $x=0$, $y^2=\frac{2}{2}=1$. ✔

Rearrange to standard form: $y^{\prime }-(\tan x)y=(\cos x)y^4.$ So $P=-\tan x$, $Q=\cos x$, $n=4$, $1-n=-3$, $v=y^{-3}$. $v^{\prime }+(-3)(-\tan x)v=(-3)\cos x⟹v^{\prime }+3\tan xv=-3\cos x.$ IF: $\mu =e^{\int 3\tan xdx}=e^{-3\ln |\cos x|}=|\cos x{|}^{-3}={\sec }^{3}x$ (on an interval where $\cos x>0$). Then $({\sec }^{3}xv{)}^{\prime }=-3\cos x{\sec }^{3}x=-3{\sec }^{2}x.$ Integrate: ${\sec }^{3}xv=-3\tan x+C\Rightarrow v={\cos }^{3}x(C-3\tan x).$ Back-substitute $v=y^{-3}$: $\boxed{y^{-3}={\cos }^{3}x(C-3\tan x)}$ Lost solution: dividing by $y^4$ assumed $y\ne 0$. Since $n=4>0$, $y\equiv 0$ satisfies the original equation ($0=0$) and is a genuine singular solution not captured by any finite $C$.

Why substitute $y=y_1+u$? For a Riccati equation, once one solution $y_1$ is known, writing $y=y_1+u$ cancels the constant clutter and leaves a Bernoulli equation in $u$ with $n=2$ (see Riccati Equations).

With $y=1+u$, $y^{\prime }=u^{\prime }$. Substitute: $u^{\prime }=1+x-(1+2x)(1+u)+x(1+u{)}^2.$ Expand the right side: $1+x-(1+2x)-(1+2x)u+x(1+2u+u^2).$ Constant terms: $1+x-1-2x+x=0.$ Good — $y_1=1$ was a solution, so they cancel. Linear-in-$u$ terms: $-(1+2x)u+2xu=-u.$ Quadratic: $x{u}^2.$ So $u^{\prime }=-u+x{u}^2$, i.e. $u^{\prime }+u=x{u}^2.$ This is Bernoulli with $P=1$, $Q=x$, $n=2$ — exactly Exercise 2.2! From there $\frac{1}{u}=1-x+C{e}^{-x}.$ Back to $y$: $u=y-1$, so $\boxed{\frac{1}{y-1}=1-x+C{e}^{-x}}$

Why flip the roles? As written, $y$-powers sit in the denominator — not the standard shape. Invert: $\frac{dx}{dy}=\frac{2x-x^{2}y}{y}.$ $\frac{dx}{dy}=\frac{2}{y}x-y{x}^2⟹\frac{dx}{dy}-\frac{2}{y}x=-y{x}^2.$ Now it is Bernoulli in $x(y)$: independent variable $y$, dependent $x$, with $P(y)=-\frac{2}{y}$, $Q(y)=-y$, $n=2$, $1-n=-1$, $v=x^{-1}$. $v^{\prime }+(-1)\left(-\frac{2}{y}\right)v=(-1)(-y)⟹\frac{dv}{dy}+\frac{2}{y}v=y.$ IF: $\mu =e^{\int 2/ydy}=e^{2\ln |y|}=y^2.$ Then $(y^{2}v{)}^{\prime }=y^3.$ Integrate: $y^{2}v=\frac{y^4}{4}+C\Rightarrow v=\frac{y^2}{4}+\frac{C}{y^2}.$ Back-substitute $v=1/x$: $\boxed{\frac{1}{x}=\frac{y^2}{4}+\frac{C}{y^2}}$

$P=\frac{a}{x}$, $Q=b{x}^k$. With $v=y^{1-n}$: $v^{\prime }+(1-n)\frac{a}{x}v=(1-n)b{x}^k.$ Let $m=(1-n)a$ and $c=(1-n)b$. IF: $\mu =e^{\int m/xdx}=x^m.$ Then $(x^{m}v{)}^{\prime }=c{x}^{k+m}.$ Generic case ($k+m\ne -1$): $x^{m}v=\frac{c{x}^{k+m+1}}{k+m+1}+C⟹v=\frac{c{x}^{k+1}}{k+m+1}+C{x}^{-m}.$ Back-substitute: $\boxed{y^{1-n}=\frac{(1-n)b{x}^{k+1}}{k+(1-n)a+1}+C{x}^{-(1-n)a}}$ Logarithmic case: if $k+m=-1$, i.e. $k+(1-n)a=-1$, the integral $\int c{x}^{-1}dx=c\ln |x|$, giving $y^{1-n}=x^{-m}(c\ln |x|+C).$

Standard form: $y^{\prime }-y=-t{y}^2$, Bernoulli with $P=-1$, $Q=-t$, $n=2$, $1-n=-1$, $v=y^{-1}$. $v^{\prime }+(-1)(-1)v=(-1)(-t)⟹v^{\prime }+v=t.$ IF: $\mu =e^t$. Then $(e^{t}v{)}^{\prime }=t{e}^t.$ Integrate by parts: $\int t{e}^{t}dt=(t-1)e^t+C.$ So $e^{t}v=(t-1)e^t+C\Rightarrow v=(t-1)+C{e}^{-t}.$ IC: $y(0)=2\Rightarrow v(0)=\frac{1}{2}$. So $\frac{1}{2}=(0-1)+C=-1+C\Rightarrow C=\frac{3}{2}.$ Thus $v=t-1+\frac{3}{2}{e}^{-t}.$ Back-substitute $v=1/y$: $\boxed{y(t)=\frac{1}{t-1+\frac{3}{2}{e}^{-t}}}$ Long-time behaviour: as $t\to \infty$, $e^{-t}\to 0$ so $v\approx t-1\to \infty$, hence $y\approx \frac{1}{t}\to 0$. The tightening carrying capacity ($-t{y}^2$ term grows) drives the population to extinction like $1/t$. See the curve below.

As $n\to 1$, $1-n\to 0$, so $v=y^{1-n}\to y^0=1$ — a constant, carrying no information. The linear ODE $v^{\prime }+(1-n)Pv=(1-n)Q$ becomes $0=0$ in the limit: both sides vanish. So the machinery collapses.

What actually happens at $n=1$: the original equation $y^{\prime }+Py=Q{y}^1$ is $y^{\prime }+(P-Q)y=0,$ which is separable / linear-homogeneous: $\frac{dy}{y}=(Q-P)dx$, giving $y=C{e}^{\int (Q-P)dx}$. No substitution required — solve directly. This is exactly why the definition excludes $n=1$ (see Separable Equations).