Question

differential equations in the form of F(x,y,p), F(y/x,p) or differential equations containing p(p=dy/dx)

1. 4y(p^2)+2xp-y=0
2. (x^2)(p^2)+3xyp+2(y^2)=0
3. (x^2)(p^2)-(2(xy-2)p)+2(y^2)=0

Answer 1

Answer on Question #79725 – Math – Differential Equation

Question

Differential equations in the form of $F(x,y,p), F(y/x,p)$ or differential equations containing $p(p = dy/dx)$

1.

$$4yp^2 + 2xp - y = 0$$

Solution

$$\begin{aligned} y(4p^2 - 1) + 2xp &= 0 \\ y &= \frac{2xp}{1 - 4p^2} \\ \frac{dy}{dx} &= p = \frac{2p}{1 - 4p^2} + 2x \frac{p'(x)(1 - 4p^2) + 8pp'(x)}{(1 - 4p^2)^2} \\ p &- \frac{2p}{1 - 4p^2} = 2 \cdot \frac{1 - 4p^2 + 8p}{(1 - 4p^2)^2} xp'(x) \\ \frac{p - 4p^3 - 2p}{1 - 4p^2} &= 2 \cdot \frac{1 - 4p^2 + 8p}{(1 - 4p^2)^2} xp'(x) \\ \frac{dx}{2x} &= \frac{4p^2 - 8p - 1}{p(1 + 4p^2)(1 - 4p^2)} dp \\ \int \frac{dx}{2x} &= \int \frac{4p^2 - 8p - 1}{p(1 + 4p^2)(1 - 2p)(1 + 2p)} dp \\ \frac{A}{p} + \frac{Bp + C}{1 + 4p^2} + \frac{D}{1 - 2p} + \frac{E}{1 + 2p} &= \frac{4p^2 - 8p - 1}{p(1 + 4p^2)(1 - 2p)(1 + 2p)} \\ A(1 + 4p^2)(1 - 4p^2) + (Bp + C)p(1 - 4p^2) + Dp(1 + 4p^2)(1 + 2p) + \\ + Ep(1 + 4p^2)(1 - 2p) &= 4p^2 - 8p - 1 \end{aligned}$$

So, we have

$$-16A - 4B + 8D - 8E = 0$$ $$-4C + 4D + 4E = 0$$ $$B + 2D - 2E = 4$$ $$C + D + E = -8$$ $$A = -1$$

Then:

$$4 - B + 2D - 2E = 0$$ $$-C + D + E = 0$$ $$B + 2D - 2E = 4$$ $$C + D + E = -8$$

Solving the given system:

$$C = -4$$ $$B = 4$$ $$\left\{ \begin{array}{c} D = E \\ D + E = -4 \end{array} \right. \Rightarrow D = E = -2$$

So far:

$$\int \frac{dx}{2x} = \int \left(- \frac{1}{p} + 4 \cdot \frac{p - 1}{1 + 4p^2} - \frac{2}{1 - 2p} - \frac{2}{1 + 2p}\right) dp$$ $$\int \frac{p - 1}{1 + 4p^2} dp = \int \frac{d(p^2)}{2(1 + 4p^2)} dp - \int \frac{dp}{1 + 4p^2} = \frac{1}{8} \ln(1 + 4p^2) - \frac{1}{2} \tan^{-1} 2p$$

Answer:

$$\ln 2x = - \ln p + \ln(1 - 2p) - \ln(1 + 2p) + \frac{1}{2} \ln(1 + 4p^2) - 2 \tan^{-1} 2p + c$$

Question

2.

$$x ^ {2} p ^ {2} + 3 x y p + 2 y ^ {2} = 0$$

Solution

$$\begin{array}{l} (x p + y) ^ {2} + x y p + y ^ {2} = 0 \\ (x p + y) ^ {2} + y (x p + y) = 0 \\ (x p + y) (x p + 2 y) = 0 \\ \end{array}$$

Case 1:

$$x p + y = 0$$ $$p = \frac {d y}{d x} = - \frac {y}{x}$$ $$\ln y = - \ln x + \ln c$$ $$x y = c$$

Case 2:

$$x p + 2 y = 0$$ $$p = \frac {d y}{d x} = - \frac {2 y}{x}$$ $$\ln y = - 2 \ln x + \ln c$$ $$x ^ {2} y = c$$

Question

3.

$$x ^ {2} p ^ {2} - 2 p (x y - 2) + 2 y ^ {2} = 0$$

Solution

$$u = y, v = x y$$ $$\frac {d v}{d x} = x p + y$$ $$P = \frac {d v}{d u} = \frac {x p + y}{p}$$ $$p = \frac {y}{P - x}$$ $$x ^ {2} p ^ {2} - 2 p x y + 4 p + 2 y ^ {2} = 0$$ $$\frac {x ^ {2} y ^ {2}}{(P - x) ^ {2}} - \frac {2 x y ^ {2}}{P - x} + \frac {4 y}{P - x} + 2 y ^ {2} = 0$$ $$x ^ {2} y ^ {2} + (4 y - 2 x y ^ {2}) (P - x) + 2 y ^ {2} (P - x) ^ {2} = 0$$ $$x ^ {2} y ^ {2} - 4 x y + 4 P y + 2 x ^ {2} y ^ {2} - 2 P x y ^ {2} + 2 x ^ {2} y ^ {2} - 4 P x y ^ {2} + 2 y ^ {2} P ^ {2} = 0$$ $$- 4 x y + 4 P y + 5 x ^ {2} y ^ {2} - 6 P x y ^ {2} + 2 y ^ {2} P ^ {2} = 0$$ $$5 v ^ {2} - 4 v + 4 P u - 6 P u v - 2 u ^ {2} P ^ {2} = 0$$

The solution of this equation cannot be determined.

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