Question

7 Obtain the differential equation associated with the given primitive
lny=Ax^2 + B, A and B being arbitrary constants.
(I) xy(d^2y/dx^2) – y dy/dx − x(dy/dx)^2=0
(II) xy (d^2y/dx^2) − 2y dy/dx=0
(III) 3xy (d^2y/dx^2) + 2y dy/dx − x(dy/dx)^2 =0
(IV) Xy (d^2y/dx^2) − x(dy/dx)^2=0
8 Solve y(xy+1)dx+x(1+xy+x^2y^2)dy=0
(I) 3x^2y^2lny−2xy−3=Cx^2y^2
(II) 2x^2y^2lny−2xy−1=Cx^2y^2
(III) 2x^2y^2lny−xy=Cx^2y^2
(IV) 2x^3y^2lny−2xy−1=Cx^2y^5

Answer 1

Answer on Question #59071 – Math – Differential Equations

Question

7 Obtain the differential equation associated with the given primitive $\ln y = Ax^2 + B$, $A$ and $B$ being arbitrary constants.

(I) $xy\frac{d^2y}{dx^2} - y\frac{dy}{dx} - x\left(\frac{dy}{dx}\right)^2 = 0$

(II) $xy\frac{d^2y}{dx^2} - 2y\frac{dy}{dx} = 0$

(III) $3xy\frac{d^2y}{dx^2} + 2y\frac{dy}{dx} - x\left(\frac{dy}{dx}\right)^2 = 0$

(IV) $xy\frac{d^2y}{dx^2} - x\left(\frac{dy}{dx}\right)^2 = 0$

Solution

If $\ln y = Ax^2 + B$, when $y = y(x)$, then

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

Differentiate both sides of this equality:

$$\frac{d}{dx}(A) = \frac{d}{dx}\left(\frac{\ln y - B}{x^2}\right),$$ $$0 = \frac{x^2 \frac{d}{dx}(\ln y - B) - (\ln y - B) \frac{d}{dx}(x^2)}{x^4}.$$ $$0 = \frac{x^2 \frac{1}{y} \frac{dy}{dx} - 2x(\ln y - B)}{x^4}$$

or

$$x^2 \frac{1}{y} \frac{dy}{dx} - 2x(\ln y - B) = 0.$$

Rewrite it as

$$x^2 \frac{1}{y} \frac{dy}{dx} = 2x(\ln y - B),$$ $$\ln y - B = \frac{x}{2y} \frac{dy}{dx}.$$

Differentiate both sides of the above equality:

$$\frac{d}{dx}(\ln y - B) = \frac{d}{dx}\left(\frac{x}{2y} \frac{dy}{dx}\right),$$ $$\frac{1}{y} \frac{dy}{dx} = \frac{y\left(\frac{dy}{dx} + x \frac{d^2y}{dx^2}\right) - x \frac{dy}{dx} \cdot \frac{dy}{dx}}{2y^2},$$ $$2y \frac{dy}{dx} = y\left(\frac{dy}{dx} + x \frac{d^2y}{dx^2}\right) - x\left(\frac{dy}{dx}\right)^2,$$

or

$$\frac{xy d^2y}{dx^2} - \frac{ydy}{dx} - x\left(\frac{dy}{dx}\right)^2 = 0$$

Answer: (I) $xy\frac{d^2y}{dx^2} - y\frac{dy}{dx} - x\left(\frac{dy}{dx}\right)^2 = 0$.

Question

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

(I) $3x^{2}y \ln y - 2xy - 3 = Cx^{2}y^{2}$

(II) $2x^{2}y^{2}\ln y - 2xy - 1 = Cx^{2}y^{2}$

(III) $2x^{2}y^{2}\ln y - xy = Cx^{2}y^{2}$

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

Solution

We may use the substitution $xy = z$, $z = z(y)$. Then $\frac{dx}{dy} = x' = \frac{z'y - z}{y^2}$

$$\frac{dx}{dy} = - \frac{x(1 + xy + x^2y^2)}{y(xy + 1)} = - \frac{z(1 + z + z^2)}{y^2(1 + z)} = \frac{z'y - z}{y^2}$$

or

$$z'y = z - \frac{z(1 + z + z^2)}{(1 + z)} = - \frac{z^3}{1 + z}.$$

We have

$$- \frac{(1 + z)dz}{z^3} = \frac{dy}{y}, \quad \int \frac{(1 + z)dz}{z^3} = - \frac{1}{2z^2} - \frac{1}{z}, \quad \int \frac{dy}{y} = \ln y.$$

So the solutions of the differential equation are

$$\frac{1}{2x^2y^2} + \frac{1}{xy} = \ln y + C$$

or

$$2x^2y^2 \ln y - 2xy - 1 = Cx^2y^2$$

Answer: (II) $2x^{2}y^{2}\ln y - 2xy - 1 = Cx^{2}y^{2}$

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