Question

Find the general solution dydx=2y2+3xy/x2

Answer 1

Answer on Question #69630 – Math – Differential Equations

Question

1. Find the general solution of $\frac{dy}{dx} = (2y^2 + 3xy) / x^2$ .

Solution

$$\frac{dy}{dx} = \frac{2y^2 + 3xy}{x^2} \Rightarrow y' = \frac{3}{x}y + \frac{2}{x^2}y^2 \Rightarrow$$ $$y' - \frac{3}{x}y = \frac{2}{x^2}y^2. \quad (1)$$

Method 1

Equation (1) is a first-order linear differential equation.

1. Let's make the substitution: $y = UV$

2. Then $y' = U'V + UV'$ .

3. Let's put the substitution into the differential equation (1):

$$U'V + UV' - \frac{3}{x}UV = 2x^{-2}U^2V^2.$$ $$U'V + U\left(V' - \frac{3}{x}V\right) = 2x^{-2}U^2V^2.$$ $$\left\{ \begin{array}{c} V' - \frac{3}{x}V = 0 \\ U'V = 2x^{-2}U^2V^2 \end{array} \right. \quad (2)$$

4. Let's solve the first differential equation of the system (2):

$$V' - \frac{3}{x}V = 0.$$ $$\frac{dV}{dx} = \frac{3}{x}V.$$ $$\frac{dV}{V} = \frac{3}{x}dx, \quad V \neq 0$$ $$\int \frac{dV}{V} = \int \frac{3}{x}dx.$$ $$\ln|V| = 3\ln|x| + C.$$ $$V = Cx^3.$$

Put $C = 1$ , then

$$V = x^3.$$

5. Let's solve the second differential equation of the system (2):

$$U'V = 2x^{-2}U^2V^2 \Rightarrow U' = 2x^{-2}U^2V \Rightarrow U' = 2x^{-2}U^2x^3 \Rightarrow U' = 2xU^2.$$

Then

$$\frac{dU}{U^2} = 2xdx, \quad U \neq 0 \Rightarrow \int \frac{dU}{U^2} = \int 2xdx \Rightarrow -\frac{1}{U} = x^2 + C \Rightarrow U = -\frac{1}{x^2 + C}$$

6. The general solution of the differential equation (1):

$$y = UV = \left(-\frac{1}{x^2 + C}\right)x^3 = -\frac{x^3}{x^2 + C}.$$

If $U = 0$ or $V = 0$ then $y = 0$ is the special solution of the differential equation (1).

Method 2

The equation (1) is homogeneous first-order differential equation because $\frac{3\lambda y}{\lambda x} = \frac{3y}{x}$ and

$$2 \frac {(\lambda y) ^ {2}}{(\lambda x) ^ {2}} = 2 \frac {y ^ {2}}{x ^ {2}}.$$

1. Let's make the substitution $y = ux$, where $u$ is a new variable,

2. then $y' = (ux)' = u'x + u$.

3. Let's put the substitution into the differential equation (1):

$$u ^ {\prime} x + u - 3 u = 2 u ^ {2},$$ $$x \frac {d u}{d x} = 2 u ^ {2} + 2 u,$$ $$x \frac {d u}{d x} = 2 u (u + 1),$$ $$\frac {d u}{u (u + 1)} = 2 \frac {d x}{x}, u \neq 0, u \neq - 1$$ $$\left(\frac {1}{u} - \frac {1}{u + 1}\right) d u = 2 \frac {d x}{x},$$ $$\int \left(\frac {1}{u} - \frac {1}{u + 1}\right) d u = 2 \int \frac {d x}{x},$$ $$\ln | u | - \ln | u + 1 | = 2 \ln x + \ln | C _ {1} |,$$ $$\ln \frac {| u |}{| u + 1 |} = \ln | C _ {1} | x ^ {2}$$ $$\frac {u}{u + 1} = C _ {2} x ^ {2},$$ $$\frac {u + 1}{u + 1} - \frac {1}{u + 1} = C _ {2} x ^ {2},$$ $$\frac {1}{u + 1} = 1 - C _ {2} x ^ {2},$$ $$u + 1 = \frac {1}{1 - C _ {2} x ^ {2}}$$ $$u = \frac {1}{1 - C _ {2} x ^ {2}} - 1$$ $$u = \frac {1 - 1 + C _ {2} x ^ {2}}{1 - C _ {2} x ^ {2}},$$ $$u = - \frac {x ^ {2}}{x ^ {2} - \frac {1}{C _ {2}}}.$$

4. The general solution of the differential equation (1):

$$\frac {y}{x} = - \frac {x ^ {2}}{x ^ {2} + C},$$ $$y = - \frac {x ^ {3}}{x ^ {2} + C}.$$

If $u = 0$ , then $y = 0$ is the special solution of the differential equation (1).

If $u = -1$ , then $y = -x$ is a solution of the differential equation (1) which is a particular case of (3) if we put $C = 0$ .

Answer:

$y = -\frac{x^3}{x^2 + C}$ is the general solution;

$y = 0$ is the special solution.

Answer provided by https://www.AssignmentExpert.com