Question

dy/dx = - (4+xy/x^2) + y^2 , find a solution to the ordinary differential equation in the form y1= ax^-1

Answer 1

Answer on Question #56726 – Math – Differential Equations

Question

$$\frac{dy}{dx} = -\frac{4 + xy}{x^2} + y^2.$$

Find a solution to the ordinary differential equation in the form

$$y = \frac{a}{x}$$

Solution

$$y' = -\frac{4}{x^2} - \frac{y}{x} + y^2.$$

Let $y_1 = \frac{a}{x}$ be a solution of equation. Then

$$-\frac{a}{x^2} = -\frac{4}{x^2} - \frac{a}{x^2} + \frac{a^2}{x^2},$$ $$a^2 = 4,$$

hence $a = \pm 2$.

Case 1

Let $a = 2$.

Substituting $y = z + \frac{2}{x}$ we obtain

$$z' - \frac{2}{x^2} = z^2 + \frac{4z}{x} + \frac{4}{x^2} - \frac{z}{x} - \frac{2}{x^2} - \frac{4}{x^2};$$ $$z' - \frac{3z}{x} = z^2;$$ $$\frac{z'}{z^2} - \frac{3}{zx} = 1.$$

Substituting $t = \frac{1}{z}$, $t' = -\frac{1}{z^2}z'$.

$$t' + \frac{3t}{x} = -1$$

First we solve $\hat{t} + \frac{3\hat{t}}{x} = 0$;

$$\frac{d\hat{t}}{\hat{t}} = -\frac{3dx}{x}; \ln\hat{t} = -3\ln x + \ln C; \hat{t} = \frac{C}{x^3}.$$

Let $t = \frac{j(x)}{x^3}$, then $\frac{j'(x)x^3 - j(x)3x^2}{x^6} + \frac{3j(x)}{x^4} = -1$.

Hence $j'(x) = -x^3$, which gives $j(x) = -\frac{x^4}{4} + C_1$, where $C_1$ is an arbitrary real constant.

Thus, $t = -\frac{x}{4} + \frac{C_1}{x^3} = \frac{-x^4 + 4C_1}{4x^3} = \frac{x^4 - 4C_1}{-4x^3} = \frac{x^4 + C}{-4x^3}$.

Get back to substitution $t = \frac{1}{z}$ and obtain $z = \frac{-4x^3}{x^4 + C}$.

Since $y = z + \frac{2}{x}$, then $y = \frac{-4x^3}{x^4 + C} + \frac{2}{x} = \frac{2C - 2x^4}{x(x^4 + C)}$, where $C$ is an arbitrary real constant.

Case 2

Let $a = -2$.

Substituting $y = z - \frac{2}{x}$ we obtain

$$z' + \frac{2}{x^2} = z^2 - \frac{4z}{x} + \frac{4}{x^2} - \frac{z}{x} + \frac{2}{x^2} - \frac{4}{x^2}$$ $$z' = z^2 - \frac{5z}{x}$$ $$\frac{z'}{z^2} + \frac{5}{zx} = 1.$$

Substituting $t = \frac{1}{z}$, $t' = -\frac{1}{z^2} z'$.

$$t' + \frac{5t}{x} = 1$$

First we solve

$$\hat{t}' + \frac{5\hat{t}}{x} = 0$$ $$\frac{d\hat{t}}{t} = -\frac{5dx}{x},$$ $$\ln \hat{t} = -5 \ln x + \ln C,$$ $$\hat{t} = \frac{C}{x^5}.$$

Let $t = \frac{j(x)}{x^5}$, then $\frac{j'(x)x^5 - j(x)5x^4}{x^{10}} + \frac{5j(x)}{x^6} = 1$.

Hence $j'(x) = x^5$, which gives $j(x) = \frac{x^6}{6} + C_1$, where $C_1$ is an arbitrary real constant, thus, $t = \frac{x}{6} + \frac{C_1}{x^5} = \frac{x^6 + 6C_1}{6x^5} = \frac{x^6 + C}{6x^5}$.

Get back to substitution $t = \frac{1}{z}$ and obtain

$$z = \frac{6x^5}{x^6 + C}.$$

Since $y = z - \frac{2}{x}$, then $y = \frac{6x^5}{x^6 + C} - \frac{2}{x} = \frac{6x^6 - 2x^6 - 2C}{x(x^6 + C)} = \frac{4x^6 - 2C}{x(x^6 + C)}$, where $C$ is an arbitrary real constant.

Answer: $y_1 = \frac{2C_1 - 2x^4}{x(x^4 + C_1)}$, $y_2 = \frac{4x^6 - 2C_2}{x(x^6 + C_2)}$, where $C_1, C_2$ are arbitrary real constants.

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