Tag: Integral Curves

Plot the isoclines of the equation y = xy′ + y′′

by RoRi

January 31, 2016

If $y$ is the solution of a differential equation, the points at which $y'$ has a constant value $C$ lie on a line for each $C$ . This line is called an isocline.

Plot some isoclines and construct a direction field of the equation

$y = x \frac{dy}{dx} + \left( \frac{dy}{dx} \right)^2.$

Determine a one-parameter family of solutions of this equation from the appearance of the direction field.

Incomplete.

Show that the isoclines of y′ = x + y form a one-parameter family of lines

by RoRi

January 31, 2016

If $y$ is the solution of a differential equation, the points at which $y'$ has a constant value $C$ lie on a line for each $C$ . This line is called an isocline.

Show that the isoclines of the differential equation

$y' = x + y$

form a one-parameter family of straight lines. Make a plot of the isoclines corresponding to the slopes $0, \ \pm \frac{1}{2}, \ \pm 1, \ \pm \frac{3}{2}, \ \pm 2$ . Using these isoclines, construct a direction field and sketch the integral curve passing through the origin. Identify one of the integral curves is also an isocline.

The isoclines of $y' = x + y$ are the curves $x + y= C$ which implies $y = -x + C$ . These are straight lines with slope $-1$ . The integral curve passing through the origin is $y = -x$ . The isocline $x+y=-1$ is also an integral curve.

Plot the isoclines of the differential equation y′ = x² + y²

by RoRi

January 31, 2016

If $y$ is the solution of a differential equation, the points at which $y'$ has a constant value $C$ lie on a line for each $C$ . This line is called an isocline.

For the differential equation

$y' = x^2 + y^2$

plot the isoclines corresponding to the constant slopes $\frac{1}{2}, \ 1, \ \frac{3}{2}$ , and 2. Using these isoclines, construct a direction field for the equation and determine the shape of the integral curve which passes through the origin.

The isoclines of the differential equation $y' = x^2 + y^2$ are concentric circles centered at the origin with slope equal to the radius of the circle. The shape of the integral curve passing through the origin is cubic.

Find a first-order differential equation whose integral curves are all circles through (1,1) and (-1,-1)

by RoRi

January 31, 2016

Find a first-order differential equation whose integral curves consist of all circles through the points $(1,1)$ and $(-1,-1)$ .

Circles going through both the points $(1,1)$ and $(-1,-1)$ must have their center on the line $y = -x$ , say at $(C,-C)$ . The radius is given by

$r^2 = (C-1)^2 +(-C-1)^2 = 2(C^2+1).$

Therefore we have the equation

$(x-C)^2 + (y+C)^2 = 2(C^2+1).$

Differentiating both sides with respect to $x$ ,

$(x-C)^2 + (y+C)^2 = 2(C^2 + 1) \quad \implies \quad 2(x-C) + 2(y+C)y' = 0.$

From the original equation we can solve for the constant,

$(x-C)^2 + (y+C)^2 = 2(C^2 + 1) \quad \implies \quad C = \frac{2-x^2-y^2}{2y-2x}.$

Therefore,

$\begin{align*} && 2(x-C) + 2(y+C)y' &= 0 \\ \implies && x - \frac{2-x^2-y^2}{2y-2x} + y'y + \left( \frac{2-x^2-y^2}{2y-2x} \right) y' &= 0 \\ \implies && 2xy - 2x^2 - 2 + x^2 + y^2 + 2y^2 y' - 2xyy' + 2y' - x^2 y' -y^2y' &= 0\\ \implies && y' (x^2 - y^2 + 2xy-2) - y^2 - 2xy + x^2 + 2 &= 0. \end{align*}$

Find a first-order differential equation having all circles through (1,0) and (-1,0) as integral curves

by RoRi

January 31, 2016

Find a first-order differential equation with all circles through the points $(1,0)$ and $(-1,0)$ as integral curves.

Circles that go through both the points $(1,0)$ and $(-1,0$ must have center on the $y$ -axis, at $(0,C)$ say. Then the radius is given by

$r = \sqrt{1+C^2}.$

Therefore, they all satisfy the equation

$x^2 + (y-C)^2 = 1+C^2.$

Differentiating both sides of this equation with respect to $x$ we have,

$x^2 + (y-C)^2 = 1+C^2 \quad \implies \quad 2x + 2(y-C)y' = 0.$

From the original equation we can also solve for the constant,

$x^2 + (y-C)^2 = 1+C^2 \quad \implies \quad C = \frac{x^2 + y^2 - 1}{2y}.$

Therefore we have,

$\begin{align*} && 2x + 2(y-c)y' &= 0 \\ \implies && x + yy' - \left( \frac{x^2+y^2-1}{2y} \right)y' &= 0\\ \implies && 2xy + 2y^2 y' - (x^2+y^2-1)y' &= 0\\ \implies && (y^2 - x^2 + 1)y' + 2xy &= 0 \\ \implies && (x^2-y^2-1)y' - 2xy &= 0. \end{align*}$

Find a first-order differential equation having the family arctan y + arcsin x = C as integral curves

by RoRi

January 31, 2016

Find a first-order differential equation having the family

$\arctan y + \arcsin x = C$

as integral curves.

We differentiate both sides of the given equation with respect to $x$ ,

$\begin{align*} &&\arctan y + \arcsin x &= C \\ \implies && \left( \frac{1}{1+y^2} \right) y' + \frac{1}{\sqrt{1-x^2}} &= 0 \\ \implies && \big( \sqrt{1-x^2} \big) y' + y^2 + 1 &= 0. \end{align*}$

This is a first-order differential equation with the given family of curves as integral curves.

Find a first-order differential equation having the family y = c (cos x) as integral curves

by RoRi

January 31, 2016

Find a first-order differential equation having the family

$y = C \cdot \cos x$

as integral curves.

First, we differentiate both sides of the given equation with respect to $x$ ,

$y = C \cdot \cos x \quad \implies \quad y' = -C \cdot \sin x.$

From the original equation we can also solve for the constant,

$y = C \cdot \cos x \quad \implies quad C = \frac{y}{\cos x}.$

Therefore,

$y' = -C \cdot \sin x \quad \implies \quad y' = -y \tan x \quad \implies quad y' + y \tan x = 0.$

This is a first-order differential equation with the given family of curves as integral curves.

Find a first-order differential equation having the family y⁴ (x + 2) = C(x – 2) as integral curves

by RoRi

January 31, 2016

Find a first-order differential equation having the family

$y^4 (x+2) = C(x-2)$

as integral curves.

First, we differentiate both sides of the given equation with respect to $x$ ,

$y^4 (x+2) = C(x-2) \quad \implies \quad 4y^3 y' (x+2) + y^4 = C.$

From the original equation we can solve for the constant,

$y^4 (x+2) = C(x-2) \quad \implies \quad C = \frac{y^4(x+2)}{x-2}.$

Therefore we have,

$\begin{align*} &&4y^3 y' (x+2) + y^4 &= C \\ \implies && 4y^3 y' (x+2) + y^4 &= \frac{y^4(x+2)}{x-2} \\ \implies && 4y^3 y' (x^2-4) + y^4(x-2) &= y^4(x+2) \\ \implies && 4y^3 (x^2-4) y' - 4y^4 &= 0 \\ \implies && (x^2-4) y' - y &= 0. \end{align*}$

This is a first-order differential equation with the given family of curves as integral curves.

Find a first-order differential equation having the family y = C(x-1)e^x as integral curves

by RoRi

January 31, 2016

Find a first-order differential equation having the family

$y = C(x-1)e^x$

as integral curves.

Starting with the given equation we differentiate both sides with respect to $x$ ,

$y = C(x-1)e^x \quad \implies \quad y' = Ce^x + C(x-1)e^x = Ce^x + y.$

From the original equation we can solve for $C$ ,

$y = C(x-1)e^x \quad \implies \quad C = \frac{ye^{-x}}{x-1}.$

Therefore, we have the first-order differential equation

$y' = \frac{y}{x-1} + y \quad \implies \quad (x-1)y' - xy = 0,$

having the given family as integral curves.

Find a first-order differential equation having the family x² + y² + 2Cy = 1 as integral curves

by RoRi

January 31, 2016

Find a first-order differential equation having the family

$x^2 + y^2 + 2Cy = 1$

as integral curves.

First, we differentiate both sides of the given equation with respect to $x$ ,

$x^2 + y^2 + 2Cy = 1 \quad \implies \quad 2x + 2yy' + 2Cy' = 0.$

From the given we equation we can solve for the constant,

$x^2 + y^2 + 2Cy = 1 \quad \implies \quad C = \frac{1-x^2-y^2}{2y}.$

Therefore we have

$\begin{align*} &&2x + 2yy' + 2Cy' &= 0 \\ \implies && x + yy' + y' \left( \frac{1-x^2-y^2}{2y} \right) &= 0 \\ \implies && y' (x^2-y^2-1) - 2xy &= 0. \end{align*}$

This is a first-order differential equation have the given family as integral curves.