Question

8 Solve: y(xy+1) dx + x(1+xy+x^2y^2) dy=0
(a) 2x^2y^2lny − 2xy −1=Cx^2y^2
(b) 3x^2y^2 lny − 2xy −3=Cx^2y^2
© 2x^2y^2 lny − xy=Cx^2y^2
(d) 2x^3y^2lny −2xy −1=Cx^2y^5
9 Solve
Xdy – ydx –√ x^2−y^2 dx = 0
(a) Cx = 2e^arcsiny/x
(b) Cx = e^arcsiny/x
© Cx = e^arcsin2y/3x
(d) Cx = e^arccosy/x
10 The population of student P at NOUN increases at a rate proportional to the population and to the addition of 150,250 and the population divided by 3, the differential equation of this statement is
(a) dP/dT=3 kP(150,250+P)/4
(b) dP/dT = 2kP(150,250+P)/3
© 5 dP/dT = kP(150,250+P)/3
(d) dP/dT=kP(150,250+P)/3

Answer 1

ANSWER ON QUESTION #59047 - MATH - DIFFERENTIAL EQUATIONS

QUESTION 1

Solve $y(xy + 1)dx + x(1 + xy + x^2y^2)dy = 0$

a) $2x^{2}y^{2}\ln |y| - 2xy - 1 = Cx^{2}y^{2}$

b) $3x^{2}y^{2}\ln |y| - 2xy - 3 = Cx^{2}y^{2}$

c) $2x^{2}y^{2}\ln |y| - xy = Cx^{2}y^{2}$

d) $2x^{3}y^{2}\ln |y| - 2xy - 1 = Cx^{2}y^{5}$

SOLUTION

In the equation, we introduce a new unknown function $y(x) = \frac{z(x)}{x}$

$$\frac{dy}{dx} = \frac{1}{x} \frac{dz}{dx} - \frac{z}{x^2}$$ $$y(xy + 1)dx + x(1 + xy + x^2y^2)dy = 0 \Leftrightarrow$$ $$y(xy + 1) + x(1 + xy + x^2y^2) \frac{dy}{dx} = 0 \Leftrightarrow$$ $$\frac{z}{x} \left(x \frac{z}{x} + 1\right) + x \left(1 + x \frac{z}{x} + x^2 \frac{z^2}{x^2}\right) \left(\frac{1}{x} \frac{dz}{dx} - \frac{z}{x^2}\right) = 0 \Leftrightarrow$$ $$\frac{z}{x}(z + 1) + \frac{dz}{dx}(1 + z + z^2) - x \frac{z}{x^2}(1 + z + z^2) = 0 \Leftrightarrow$$ $$\frac{z}{x}(1 + z - 1 - z - z^2) = -\frac{dz}{dx}(1 + z + z^2) \Leftrightarrow -\frac{z^3}{x} = -\frac{dz}{dx}(1 + z + z^2) \Leftrightarrow$$ $$\frac{dx}{x} = dz \frac{1 + z + z^2}{z^3} = dz \left(\frac{1}{z^3} + \frac{1}{z^2} + \frac{1}{z}\right)$$ $$\ln |x| + \text{CONST} = -\frac{1}{2z^2} - \frac{1}{z} + \ln |z| \Leftrightarrow \text{CONST} = -\frac{1}{2x^2y^2} - \frac{1}{xy} + \ln |xy| - \ln |x|$$ $$\text{CONST} * 2x^2y^2 = -1 - 2xy + 2x^2y^2 \ln |y| \Leftrightarrow 2x^2y^2 \ln |y| - 2xy - 1 = \underbrace{2 * \text{CONST}}_{C} * x^2y^2$$

ANSWER:

a) $2x^{2}y^{2}\ln |y| - 2xy - 1 = Cx^{2}y^{2}$

QUESTION 2

Solve $xdy - ydx - \sqrt{x^2 - y^2} dx = 0$

a) $Cx = 2e^{arcsin\frac{y}{x}}$

b) $Cx = e^{arcsin\frac{y}{x}}$

c) $Cx = e^{arcsin\frac{2y}{3x}}$

d) $Cx = e^{arccos\frac{y}{x}}$

SOLUTION

In the equation, we introduce a new unknown function $y(x) = x * z(x)$

$$\frac{dy}{dx} = z + x \frac{dz}{dx}$$ $$xdy - ydx - \sqrt{x^2 - y^2} dx = 0 \Leftrightarrow x \frac{dy}{dx} - y - \sqrt{x^2 - y^2} = 0$$ $$x\left(z + x \frac{dz}{dx}\right) - zx - \sqrt{x^2 - x^2 z^2} = 0 \Leftrightarrow xz + x^2 \frac{dz}{dx} - zx - x\sqrt{1 - z^2} = 0$$ $$x^2 \frac{dz}{dx} = x\sqrt{1 - z^2} \Leftrightarrow \frac{dz}{\sqrt{1 - z^2}} = \frac{dx}{x} \Leftrightarrow \arcsin(z(x)) = \ln|x| + \ln|C|$$ $$\ln|Cx| = \arcsin\left(\frac{y}{x}\right) \Leftrightarrow Cx = e^{arcsin\left(\frac{y}{x}\right)}$$

ANSWER:

b) $Cx = e^{arcsin\left(\frac{y}{x}\right)}$

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QUESTION 3

The population of student P at NOUN increases at a rate proportional to the population and to the addition of 150,250 and the population divided by 3, the differential equation of this statement is

a) $\frac{dP}{dT} = 3kP\frac{(150,250 + P)}{4}$

b) $\frac{dP}{dT} = 2kP\frac{(150,250 + P)}{3}$

c) $5\frac{dP}{dT} = kP\frac{(150,250 + P)}{3}$

d) $\frac{dP}{dT} = 3kP\frac{(150,250 + P)}{4}$

SOLUTION

The population of student P at NOUN increases at the population divided by 3. Its means that

$$\frac{dP}{dT} \sim \frac{P}{3}$$

The population of student P at NOUN increases at a rate proportional to the population and to the addition of 150,250 – its mean that

$$\frac{dP}{dT} \sim P + 150,250$$

THEN,

$$\frac{dP}{dT} \sim \frac{P}{3} (P + 150,250) \Leftrightarrow \frac{dP}{dT} = \underbrace{2k}_{\text{the proportionality factor}} \frac{P}{3} (P + 150,250).$$

ANSWER:

b) $\frac{dP}{dT} = 2kP\frac{(150,250 + P)}{3}$