Tag: Nonlinear Differential Equations

Find solutions of the Riccati equation y′ + y + y² = 2

by RoRi

January 31, 2016

$y' + y + y^2 = 2.$

This equation has two constant solutions. Starting with these and the previous exercise (linked above) find further solutions in the cases:

If $-2 \leq b < 1$ , find a solution on the interval $(-\infty, +\infty)$ with $y= b$ when $x = 0$ .
If $b \geq 1$ or $b<-2$ , find a solution on the interval $(-\infty, +\infty)$ with $y = b$ when $x = 0$ .

First, we find the two constant solutions. If $u$ is constant then $u' = 0$ so

$u' + u + u^2 = 2 \quad \implies \quad u^2+u-2 = 0 \quad \implies \quad u= 1 \text{ or } -2.$

From Exercise 19 (linked above) we know we can obtain additional solutions $y$ to the Riccati equation by

$y = u + \frac{1}{v}$

where $v$ is a solution of

$v' - v (P(x) + 2Q(x)u) = Q(x).$

In the present case we have $P(x) = Q(x) = 1$ , so $v$ is the solution of either

$v' -3v = 1 \qquad \text{or} \qquad v' + 3v = 1,$

for the cases $u = 1$ and $u=-2$ , respectively. Each of these is an first-order linear differential equation which we can solve using Theorem 8.3 (page 310 of Apostol). For the first one we have $P(x) = 3$ , $Q(x) = 1$ and let $a = 0$ , and $c = v(0)$ . Then we have

$\begin{align*} v &= ce^{3x} + e^{3x} \int_0^x e^{-3t} \, dt \\ &= ce^{3x} + e^{3x} \frac{1}{3}( -e^{-3x} + 1) \\ &= ce^{3x} - \frac{1}{3} + \frac{1}{3}e^{3x} \\ &= \frac{e^{3x}(3c+1) -1}{3}. \end{align*}$

This gives us the first solution

$y_1 = u + \frac{1}{v} = \frac{e^{3x}(3c+1) + 2}{e^{3x}(3c+1) - 1}.$

Evaluating the second differential equation, this time with $P(x) = -3$ , $Q(x) = 1$ , $a = 0$ and $c = v(0)$ we have,

$\begin{align*} v &= ce^{-3x} + e^{-3x} \int_0^x e^{3t} \, dt \\ &= ce^{-3x} + \frac{1}{3} - \frac{1}{3}e^{-3x} \\ &= \frac{3c-1+e^{3x}}{3e^{3x}}. \end{align*}$

Therefore, the second solution is

$y_2 = u + \frac{1}{v} = \frac{-2(3c-1) + e^{3x}}{(3c-1)+e^{3x}}.$

Finally, for the specific cases in (a) and (b).

We want $-2 \leq b < 1$ , so we choose $y_2$ . Then $y_2(0) = b \implies c = \frac{1}{b+2}$ . (This follows since $y(0) = -2 + \frac{1}{v(0)}$ .) Therefore, $(3c-1) = -\frac{b-1}{b+2}$ , so,
$\begin{align*} y_2 &= \frac{2 \left( \frac{b-1}{b+2} \right) + e^{3x}}{-\left( \frac{b-1}{b+2} \right) +e^{3x}} \\[9pt] &= \frac{ \left( \frac{b+2}{b-1} e^{3x} + 2}{\left( \frac{b+2}{b-1} \right) e^{3x} -1} \\[9pt] &= \frac{Ce^{3x} +2}{Ce^{3x} -1} \end{align*}$

where $C = \frac{b-1}{b+2}$ .
In this case we want $b \geq 1$ or $b <-2$ , so we choose $y_1$ . Since $y_1(0) = b \implies c =\frac{1}{b-1}$ we have $(3c+1) = \frac{b+2}{b-1}$ . Therefore,
$\begin{align*} y_1 &= \frac{e^{3x} \left( \frac{b+2}{b-1} \right) + 2}{e^{3x} \left( \frac{b+2}{b-1} \right) -1} \\[9pt] &= \frac{e^{3x} + 2 \left(\frac{b-1}{b+2}\right)}{e^{3x} - \left( \frac{b-1}{b+2} \right)} \\[9pt] &= \frac{e^{3x} + 2C}{e^{3x} - C} \end{align*}$

where $C = \frac{b-1}{b+2}$ .

Prove a formula for additional solutions of a Riccati equation given one solution

by RoRi

January 30, 2016

Consider a differential equation

$y' + P(x) y + Q(x) y^2 = R(x)$

called a Riccati equation. Prove that if $u$ is a solution of this equation then there are additional solutions

$y = u + \frac{1}{v}$

where $v$ satisfies a first-order linear differential equation.

Proof. Let $u$ be a function satisfying

$u' + P(x) u + Q(x)u^2 = R(x).$

Further, let $y = u + \frac{1}{v}$ where $v$ satisfies a first-order linear differential equation. Then,

$y = u + \frac{1}{v} \quad \implies \quad y' = u' - \frac{v'}{v^2}.$

Therefore,

$\begin{align*} y' + P(x)y + Q(x) y^2 &= \left( u' - \frac{v'}{v^2} \right) + \left( u + \frac{1}{v} \right)P(x) + \left( u + \frac{1}{v} \right)^2 Q(x) \\[9pt] &= u' - \frav{v'}{v^2} + P(x) u + P(x) \frac{1}{v} + Q(x) u^2 + 2Q(x) \frac{u}{v} + Q(x) \frac{1}{v^2} \\[9pt] &= R(x) - \frac{v'}{v^2} + P(x)\frac{1}{v} + 2Q(x) \frac{u}{v} + Q(x) \frac{1}{v^2}. \end{align*}$

Thus, if

$-\frac{v'}{v^2} + P(x)v + 2Q(x) \frac{v}{u} + Q(x) \frac{1}{v^2} = 0,$

then $y = u+ \frac{1}{v}$ is a solution of our equation. But,

$\begin{align*} &&-\frac{v'}{v^2} + P(x) \frac{1}{v} + 2Q(x) \frac{u}{v} + Q(x) \frac{1}{v^2} &= 0 \\[9pt] \implies && -v' + P(x) v + 2Q(x) uv + Q(x) &= 0 \\[9pt] \implies && v' - v (P(x) + 2Q(x) u) &= Q(x). \end{align*}$

This is a first-order linear differential equation (since $u$ is a function of $x$ ). Hence, $y = u + \frac{1}{v}$ is a solution of

$y' + P(x) y + Q(x)y^2 = R(x)$

where $v$ is the solution of the first-order linear differential equation

$v' - v (P(x) + 2Q(x) u) = Q(x). \qquad \blacksquare$

Find solutions of 2xyy′ + (1+x)y² = e^x for given initial values

by RoRi

January 30, 2016

Find all solutions of the nonlinear differential equation

$2xyy' + (1+x)y^2 = e^x$

on the interval $(0, +\infty)$ satisfying the initial conditions

$y = \sqrt{e}$ when $x = 1$ ;
$y = -\sqrt{e}$ when $x = 1$ ;
a finite limit as $x \to 0$ .

previous exercise

$y = f(x)$

$I$

$y' + P(x) y = Q(x) y^n \qquad \text{on } I, \qquad \text{with } (f(a))^k = b$

if and only if $(f(x))^k = g(x)$ on $I$ where $g(x) = v$ is the unique solution of the initial-value problem

$v' + kP(x) v = kQ(x) \qquad \text{on } I, \qquad \text{with } g(a) = b.$

In the present problem we have the equation
$y' + \frac{1+x}{2x} y = \frac{e^x}{2x} y^{-1} \qquad \text{on } (0, +\infty) \qquad \text{with } y = \sqrt{e} \text{ when } x = 1.$

Therefore, to apply the previous exercise we have

$P(x) = \frac{1+x}{2x}, \quad Q(x) = \frac{e^x}{2x}, n = -1, k = 1 - n = 2.$

Hence, $f(x)$ is a solution to the given equation if and only if $(f(x))^2 = g(x)$ where $g(x)$ is the unique solution to

$v' + \frac{1+x}{x}v = \frac{e^x}{x} \qquad \text{with } g(1) = e.$

We can solve this using Theorem 8.3 (page 310 of Apostol), first computing

$A(x) = \int_1^x \frac{1+t}{t} \,dt = \log x + x - 1,$

giving us

$\begin{align*} g(x) &= e \cdot e^{-\log x - x +1} + e^{-\log x - x +1} \int_1^x \frac{e^t}{t} e^{\log t + t - 1} \, dt \\[9pt] &= \frac{e^2}{xe^x} + \frac{e}{xe^x} \int_1^x e^{2t-1} \, dt \\[9pt] &= \frac{e^2}{xe^x} + \frac{1}{xe^x} \int_1^x e^{2t} \, dt\\[9pt] &= \frac{e^2}{xe^x} + \frac{1}{xe^x} \left( \frac{1}{2} e^{2t} \Bigr \rvert_1^x \right) \\[9pt] &= \frac{e^2}{xe^x} + \frac{1}{xe^x} \left( \frac{e^{2x}}{2} - \frac{e^2}{2} \right) \\[9pt] &= \frac{e^x + e^2 - x}{2x} \end{align*}$

Therefore,

$(f(x))^2 = g(x) \quad \implies \quad f(x) = \left( \frac{e^x+e^{2-x}}{2x} \right)^{\frac{1}{2}}.$
From part (a) we have $g(x) = (f(x))^2$ which implies $f(x) = \pm (g(x))^{\frac{1}{2}}$ . Since $f(1) = -\sqrt{e}$ we then have
$f(x) = - \left( \frac{e^x + e^{2-x}}{2x} \right)^{\frac{1}{2}}.$
Let $\lim_{x \to 0} f(x) = b$ where $b$ is some finite real number. Then, (using the calculations we already completed in part (a) and just changing the initial-values $a$ and $b$ )
$\begin{align*} v &= be^{-x-\log x + a + \log a} + e^{-x - \log x + a + \log a} \int_a^x \let( \frac{e^t}{t} \right) \left( e^{t + \log t - a - \log a} \right) \, dt \\[9pt] &=\frac{bae^a}{xe^x} + \left( \frac{1}{xe^x} \right) \int_a^x e^{2t} \, dt \\[9pt] &= \frac{bae^a}{xe^x} + \frac{e^{2x} - e^{2a}}{2xe^x} \\[9pt] &= \frac{e^{2x} - c}{2xe^x}. \end{align*}$

Then,

$\lim_{x \to 0} \frac{e^{2x} - c}{2xe^x} = L \quad \implies \quad \lim_{x \to 0} (e^{2x} - c) = 0 \quad \implies \quadd c = 1.$

Therefore,

$y^2 = \frac{e^{2x}-1}{2xe^x} = \frac{e^x - e^{-x}}{2x} = \frac{\sinh x}{x}.$

Find all solutions of xy′ + y = y²x² log x for given initial values

by RoRi

January 30, 2016

Find all solutions of the nonlinear differential equation

$xy' + y = y^2 x^2 \log x$

on the interval $(0, +\infty)$ satisfying the initial conditions

$y = \frac{1}{2} \qquad \text{when} \qquad x = 1.$

From a previous exercise (Section 8.5, Exercise #13) we know that a function $y = f(x)$ which is never zero on an interval $I$ is a solution of the initial value problem

$y' + P(x) y = Q(x) y^n \qquad \text{on } I, \qquad \text{with } (f(a))^k = b$

if and only if $(f(x))^k = g(x)$ on $I$ where $g(x) = v$ is the unique solution of the initial-value problem

$v' + kP(x) v = kQ(x) \qquad \text{on } I, \qquad \text{with } g(a) = b.$

In the present problem we have the equation

$y' + \frac{1}{x}y = y^2 x \log x \qquad \text{on } (0, +\infty) \qquad \text{with } y = \frac{1}{2} \text{ when } x = 1.$

Therefore, to apply the previous exercise we have

$P(x) = \frac{1}{x}, \quad Q(x) = x \log x, n = 2, k = 1 - n = -1.$

Hence, $f(x)$ is a solution to the given equation if and only if $(f(x))^{-1} = g(x)$ where $g(x)$ is the unique solution to

$v' - \frac{1}{x} v = -x \log x \qquad \text{with } g(1) = 2.$

We can solve this using Theorem 8.3 (page 310 of Apostol), first computing

$A(x) = \int_1^x -\frac{1}{t} \, dt = - \log x$

giving us

$\begin{align*} g(x) &= 2e^{\log x} + e^{\log x} \int_1^x (-t \log t) \cdot e^{-\log t} \, dt \\ &= 2x - x \int_1^x \log t \, dt \\ &= 2x - x (x \log x - x + 1) \\ &= x^2 + x - x^2 \log x. \end{align*}$

Therefore,

$(f(x))^{-1} = g(x) \quad \implies \quad f(x) = \frac{1}{x^2+x-x^2 \log x}.$

Find solutions of xy′ – 2y = 4x³y^1/2 for given initial values

by RoRi

January 30, 2016

Find all solutions of the nonlinear differential equation

$xy' - 2y = 4x^3y^{\frac{1}{2}}$

on the interval $(-\infty, +\infty)$ satisfying the initial conditions

$y = 0 \qquad \text{when} \qquad x = 1.$

From a previous exercise (Section 8.5, Exercise #13) we know that a function $y = f(x)$ which is never zero on an interval $I$ is a solution of the initial value problem

$y' + P(x) y = Q(x) y^n \qquad \text{on } I, \qquad \text{with } (f(a))^k = b$

if and only if $(f(x))^k = g(x)$ on $I$ where $g(x) = v$ is the unique solution of the initial-value problem

$v' + kP(x) v = kQ(x) \qquad \text{on } I, \qquad \text{with } g(a) = b.$

In the present problem we have the equation

$xy' - 2y = 4x^3 y^{\frac{1}{2}} \qquad \text{on } (-\infty, +\infty) \qquad \text{with } y = 0 \text{ when } x = 1.$

Therefore, to apply the previous exercise we have

$P(x) = -\frac{2}{x}, \quad Q(x) = 4x^2, n = \frac{1}{2}, k = 1 - n = \frac{1}{2}.$

Hence, $f(x)$ is a solution to the given equation if and only if $(f(x))^{\frac{1}{2}} = g(x)$ where $g(x)$ is the unique solution to

$v' - \frac{1}{x}v = 2x^2 \qquad \text{with } g(1) = 0.$

We can solve this using Theorem 8.3 (page 310 of Apostol), first computing

$A(x) = \int_1^x -\frac{1}{t} \, dt = -\log x,$

giving us

$\begin{align*} g(x) &= e^{\log x} \int_1^x 2t^2 \cdot e^{-\log t} \,dt \\ &= x \int_1^x 2t \, dt \\ &= x \left( t^2 \Bigr \rvert_1^x \right) \\ &= x (x^2 - 1) \\ &= x^3 - x. \end{align*}$

Therefore,

$(f(x))^{\frac{1}{2}} = g(x) \quad \implies \quad f(x) = (x^3-x)^2.$

Find solutions of y′ -y = -y²(x² + x + 1) for given initial values

by RoRi

January 30, 2016

Find all solutions of the nonlinear differential equation

$y' - y = -y^2 (x^2+x+1)$

on the interval $(-\infty, +\infty)$ satisfying the initial conditions

$y = 1 \qquad \text{when} \qquad x = 0.$

From a previous exercise (Section 8.5, Exercise #13) we know that a function $y = f(x)$ which is never zero on an interval $I$ is a solution of the initial value problem

$y' + P(x) y = Q(x) y^n \qquad \text{on } I, \qquad \text{with } (f(a))^k = b$

if and only if $(f(x))^k = g(x)$ on $I$ where $g(x) = v$ is the unique solution of the initial-value problem

$v' + kP(x) v = kQ(x) \qquad \text{on } I, \qquad \text{with } g(a) = b.$

In the present problem we have the equation

$y' - y = -y^2(x^2+x+1) \qquad \text{on } (-\infty, +\infty) \qquad \text{with } y = 1 \text{ when } x = 0.$

Therefore, to apply the previous exercise we have

$P(x) = -1, \quad Q(x) = -(x^2+x+1), n = 2, k = 1 - n = -1.$

Hence, $f(x)$ is a solution to the given equation if and only if $(f(x))^{-1} = g(x)$ where $g(x)$ is the unique solution to

$v' + v = x^2+x+1 \qquad \text{with } g(0) = 1.$

We can solve this using Theorem 8.3 (page 310 of Apostol), first computing

$A(x) = \int_0^x \,dt = x,$

giving us

$\begin{align*} g(x) &= e^{-x} + e^{-x} \int_0^x (t^2+t+1)e^t \, dt \\ &= e^{-x} + e^{-x} \left( e^t (t^2 - t + 2) \right) \Bigr \rvert_0^x \\ &= e^{-x} + e^{-x} ( x^2 e^x - xe^x + 2e^x - 2) \\ &= e^{-x} + x^2 - x + 2 - 2e^{-x} \\ &= x^2 - x + 2 - e^{-x}. \end{align*}$

Therefore,

$(f(x))^{-1} = g(x) \quad \implies \quad f(x) = \frac{1}{x^2-x+2-e^{-x}}.$

Find solutions of y′ – 4y = 2e^xy^1/2 for given initial values

by RoRi

January 30, 2016

Find all solutions of the nonlinear differential equation

$y' - 4y = 2e^x y^{\frac{1}{2}}$

on the interval $(-\infty, +\infty)$ satisfying the initial conditions

$y = 2 \qquad \text{when} \qquad x = 0.$

From the previous exercise (Section 8.5, Exercise #13) we know that a function $y = f(x)$ which is never zero on an interval $I$ is a solution of the initial value problem

$y' + P(x) y = Q(x) y^n \qquad \text{on } I, \qquad \text{with } (f(a))^k = b$

if and only if $(f(x))^k = g(x)$ on $I$ where $g(x) = v$ is the unique solution of the initial-value problem

$v' + kP(x) v = kQ(x) \qquad \text{on } I, \qquad \text{with } g(a) = b.$

In the present problem we have the equation

$y' - 4y = 2e^x y^{\frac{1}{2}} \qquad \text{on } (-\infty, +\infty) \qquad \text{with } y = 2 \text{ when } x = 0.$

Therefore, to apply the previous exercise we have

$P(x) = -4, \quad Q(x) = 2e^x, n = \frac{1}{2}, k = 1 - n = \frac{1}{2}.$

Hence, $f(x)$ is a solution to the given equation if and only if $(f(x))^{\frac{1}{2}} = g(x)$ where $g(x)$ is the unique solution to

$v' - 2 v = e^x \qquad \text{with } g(0) = \sqrt{2}.$

We can solve this using Theorem 8.3 (page 310 of Apostol), first computing

$A(x) = \int_0^x -2 \,dt = -2x,$

giving us

$\begin{align*} g(x) &= \sqrt{2} e^{2x} + e^{2x} \int_0^x 2e^t \cdot e^{-2t} \, dt \\ &= \sqrt{2} e^{2x} + e^{2x} (-e^{-x} + 1) \\ &= \sqrt{2} e^{2x} + e^{2x} - e^x. \end{align*}$

Therefore,

$(f(x))^{\frac{1}{2}} = g(x) \quad \implies \quad f(x) = \left( \sqrt{2} e^{2x} + e^{2x} - e^x \right)^2.$

Find all solutions of a given initial value problem

by RoRi

January 30, 2016

Let the function $v = g(x)$ be the unique solution of the initial-value problem

$v' + kP(x)v = kQ(x)$

with $g(a) = b$ where $k \neq 0$ is a constant, $P(x)$ and $Q(x)$ are continuous functions on an interval $I$ , and $a \in I$ with $b$ any real number. If $n \neq 1$ and $k = n-1$ prove that the function $y = f(x)$ for which $f(x) \neq 0$ for all $x \in I$ is a solution of

$y' + P(x) y = Q(x) y^n \qquad \text{on } I, \qquad \text{with } f(a)^k = b$

if and only if $(f(x))^k = g(x)$ for all $x \in I$ .

Proof. $(\Rightarrow)$ Assume $y = f(x)$ is a solution of $y' + P(x)y = Q(x) y^n$ . Let $v = y^{1-n} = y^k$ . Then,

$v' = ky^{k-1} y' \quad \implies \quad y' = \frac{v'}{ky^{k-1}}.$

(This division is allowed since $y \neq 0$ on $I$ implies $y^{k-1} \neq 0$ on $I$ .) Therefore,

$\begin{align*} y' + P(x) y = Q(x) y^n && \implies && \frac{v'}{ky^{k-1}} + P(x) y &= Q(x) y^n \\[9pt] && \implies && \frac{v'}{ky^{k-1+n}} + P(x) y^{1-n} &= Q(x) \\[9pt] && \implies && \frac{v'}{k} + P(x) y^k &= Q(x) &(k = 1-n) \\[9pt] && \implies && v' + kP(x) v = kQ(x). \end{align*}$

Thus, if $y = f(x)$ is a solution of $y' + P(x)y = Q(x)y^n$ then $g(x) = v = y^k$ is the unique solution of

$v' + kP(x)v = kQ(x) \qquad \text{with} \qquad g(a) = (f(a))^k = b.$

$(\Leftarrow)$ Conversely, if $(f(x))^k = g(x)$ on $I$ then

$g'(x) = k(f(x))^{k-1} f'(x) \quad \implies \quad f'(x) = \frac{1}{k} g'(x) (f(x))^{1-k}.$

Therefore,

$f'(x) + P(x) f(x) = \frac{1}{k} g'(x)(f(x))^{1-k} + P(x) f(x)$

Then, since we know by hypothesis that $v = g(x)$ is the unique solution of $v' + kP(x) v = kQ(x)$ we have

$g'(x) = kQ(x) - kP(x) g(x).$

Substituting this into the above equation (and noting that our assumption that $(f(x))^k = g(x)$ implies $(f(x))^{1-k} = \frac{f(x)}{g(x)}$ ) we have

$\begin{align*} f'(x) + P(x) f(x) &= \frac{1}{k} (kQ(x) - kP(x)g(x)) (f(x))^{1-k} + P(x) f(x) \\[9pt] &= Q(x) (f(x))^{1-k} - P(x) g(x) (f(x))^{1-k} + P(x)f(x) \\[9pt] &= Q(x) (f(x))^{1-k} - P(x) g(x) \frac{f(x)}{g(x)} + P(x) f(x) \\[9pt] &= Q(x) (f(x))^{1-k} - P(x) f(x) + P(x) f(x) \\[9pt] &= Q(x) (f(x))^{1-k} \\ &= Q(x) (f(x))^n. \end{align*}$

Therefore $y = f(x)$ is a solution of $y' + P(x)y = Q(x)y^n$ with $(f(a))^k = g(a) = b. \qquad \blacksquare$