Question

3 Derive the differential equation associated with the primitive
y=Ax^3+Bx^2 + Cx + D
where A, B, C and D are arbitrary constants.

(a) D^3y/dx^2= 0
(b) d^4y/dx^4 + d^3y/dx^3 = 0
( c) d^3y/dx^3 + d^2y/dx^2 = 0
(d) d^4y/dx^4 = 0
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
(I) y=2√4+ (dx/dy)^2
(II) Y =√1+(d^2y/dx^2)^2
(III) Y =2 √1+ (dy/dx)^2
(IV) y=3 √2+(dy/dx)^2
6 Find the differential equation of all straight lines at a unit distance from the origin
(i) (x dy/dx − y)^2=1/2)^2
(II) (x dy/dx − y)^2 =1+ (dy/dx)^2
(III) (3x dy/dx − y)^2=3+(dy/dx)^2
(IV) (2x d^2y/dx^2 − y)^2=1+(dy/dx)^2

Answer 1

Answer on Question #59070 – Math – Differential Equations

Question

3. Derive the differential equation associated with the primitive $y = Ax^3 + Bx^2 + Cx + D$ where A, B, C and D are arbitrary constants.

(a) $D^3y/dx^2 = 0$

(b) $d^4y/dx^4 + d^3y/dx^3 = 0$

(c) $d^3y/dx^3 + d^2y/dx^2 = 0$

(d) $d^4y/dx^4 = 0$

Solution

$\frac{d^4}{dx^4}(Ax^3 + Bx^2 + Cx + D) = 0$, so the differential equation is $\frac{d^4y}{dx^4} = 0$.

**Answer:** (d) $d^4y/dx^4 = 0$.

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 twice the length of the arc between the ordinates

(I) $y = 2\sqrt{4} + (dx/dy)^2$

(II) $Y = \sqrt{1 + (d^2y/dx^2)^2}$

(III) $Y = 2\sqrt{1 + (dy/dx)^2}$

(IV) $y = 3\sqrt{2 + (dy/dx)^2}$

Solution

**Area:** $S = \int_{a}^{x} y(x) dx$. **Length of arc:** $L = \int_{a}^{x} \sqrt{1 + y'(x)^2} dx$.

So $\int_{a}^{x} y(x) dx = 3 \int_{a}^{x} \sqrt{1 + y(x)^2} dx \to \frac{d}{dx} \int_{a}^{x} y(x) dx = 3 \frac{d}{dx} \int_{a}^{x} \sqrt{1 + y'(x)^2} dx \to$

$$y = 3 \sqrt{1 + y'^2}.$$

**Answer:** $y = 3\sqrt{1 + y'^2}$.

Question

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

(i) $(x dy/dx - y)^2 = 1/2)^2$

(II) $(x dy/dx - y)^2 = 1 + (dy/dx)^2$

(III) $(3 \times \frac{dy}{dx} - y)^2 = 3 + \frac{dy}{dx}^2$

(IV) $(2 \times \frac{d^2y}{dx^2} - y)^2 = 1 + \frac{dy}{dx}^2$

**Solution**

Equation of the line: $y = ax + b$

Distance from the line to origin: $d = \frac{|b|}{\sqrt{a^2 + 1}} = 1 \rightarrow b = \pm \sqrt{a^2 + 1}$

So $y = ax \pm \sqrt{a^2 + 1} \rightarrow \frac{dy}{dx} = \frac{d}{dx} \left( ax \pm \sqrt{a^2 + 1} \right) \rightarrow y' = a$

and $y = y'x \pm \sqrt{y'^2 + 1} \rightarrow (y - xy')^2 = y'^2 + 1$

**Answer:** (II) $\left( \frac{x dy}{dx - y} \right)^2 = 1 + \frac{dy}{dx}^2$

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