Question

5 Derive the differential equation for the area bounded by the arc of a curve, the x- axis, and the two ordinates, one fixed and one variable, is equal to trice the length of the arc between the ordinates
6 Find the differential equation of all straight lines at a unit distance from the origin
7 Obtain the differential equation associated with the given primitive
lny=Ax^2+B
, A and B being arbitrary constants.

Answer 1

Answer on Question #58883 – Math – Differential Equations

Question

5. Derive the differential equation for the area bounded by the arc of a curve, the $x$-axis, and the two ordinates, one fixed and one variable, is equal to thrice the length of the arc between the ordinates.

Solution

The equation of area:

$$S = \int_{x_0}^{x} y(x) \, dx,$$

where $x_0$ is a fixed ordinate, $x$ is variable ordinate.

The equation of arc length:

$$L = \int_{x_0}^{x} \sqrt{1 + (y'(x))^2} \, dx$$

It is given that $S = 3L$, which is equivalent to

$$\int_{x_0}^{x} y(x) \, dx = 3 \int_{x_0}^{x} \sqrt{1 + (y'(x))^2} \, dx$$

Differentiate both sides of (1) with respect to $x$:

$$y(x) = 3 \sqrt{1 + (y'(x))^2}$$

Differentiating both sides of (2) with respect to $x$ derive the differential equation:

$$y' = \frac{6 y' y''}{2 \sqrt{1 + (y'(x))^2}}$$ $$y' = \frac{3 y' y''}{\sqrt{1 + (y'(x))^2}}$$

If $y' = 0$, then $y = C$ is a straight line, where $C$ is an arbitrary constant;

$$S = \int_{x_0}^{x} y(x) \, dx = \int_{x_0}^{x} C \, dx = C(x - x_0),$$ $$L = \int_{x_0}^{x} \sqrt{1 + (y'(x))^2} \, dx = \int_{x_0}^{x} \sqrt{1 + (C')^2} \, dx = \int_{x_0}^{x} dx = (x - x_0).$$

If $S = 3L$, then $C(x - x_0) = 3(x - x_0)$, hence $C = 3$ and $y = 3$, its differential equation is $y' = 0$.

If $y' \neq 0$, then we do not deal with a straight line, $y = y(x)$ represents a curve.

Divide both sides of (3) by $y'$:

$$\begin{array}{l} 1 = \frac{3 y''}{\sqrt{1 + (y'(x))^2}} \\ \sqrt{1 + (y'(x))^2} = 3 y'' \\ \end{array}$$

Square both sides:

$$(y')^2 + 1 = 9 (y'')^2$$

Answer: $(y')^2 + 1 = 9 (y'')^2$

Question

6. Find the differential equation of all straight lines at a unit distance from the origin.

Solution

The equation of a circle centered in the origin and the radius of 1:

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

Differentiating both sides with respect to $x$ derive the differential equation:

$$2 y y' = -2 x;$$

$y' = -\frac{x}{y}$ is the slope of the lines at distance 1 from the origin.

From $x^2 + y^2 = 1$ it follows that $y = \sqrt{1 - x^2}$ and using $y' = -\frac{x}{y}$ we come to

$$y' = -\frac{x}{\sqrt{1 - x^2}}.$$

Answer: $y' = -\frac{x}{\sqrt{1 - x^2}}$.

Question

7. Obtain the differential equation associated with the given primitive

$$\ln y = A x ^ {2} + B$$

$A$ and $B$ being arbitrary constant.

Solution

Differentiate both sides of $\ln y = Ax^2 + B$ with respect to $x$ and derive the equation:

$$\frac {y ^ {\prime}}{y} = 2 A x$$ $$\frac {y ^ {\prime}}{x y} = 2 A$$

Take the logarithm:

$$\ln y ^ {\prime} - \ln x - \ln y = \ln 2 A$$

Differentiate both sides of the previous formula with respect to $x$:

$$\frac {y ^ {\prime \prime}}{y ^ {\prime}} - \frac {1}{x} - \frac {y ^ {\prime}}{y} = 0.$$

Answer: $\frac{y^{\prime\prime}}{y^{\prime}} -\frac{1}{x} -\frac{y^{\prime}}{y} = 0.$

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