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

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

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

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)

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

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 y = c (cos x) as integral curves

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

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.