Question

Identify the following differential equations and hence solve them

1) y'=-4/x^2 -y/x +y^2
2) y= xy'+1- ln y'

Answer 1

Question #74300, Math / Differential Equations

Identify the following differential equations and hence solve them

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

2) $y = xy' + 1 - \ln y'$

1. Solution

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

This is the first-order nonlinear ordinary differential equation

$$x^2 y' = -4 - xy + x^2 y^2$$ $$x^2 y^2 = x^2 y' + xy + 4$$ $$x^2 y^2 = x(xy' + y) + 4$$ $$xy' + y = (xy')$$ $$(xy)^2 = x(xy)' + 4$$

Substitute $z = xy$

$$z^2 = xz' + 4$$ $$xz' = z^2 - 4$$ $$\frac{dz}{z^2 - 4} = \frac{dx}{x}$$ $$\int \frac{dz}{z^2 - 4} = \int \frac{dx}{x}$$ $$\frac{1}{4} \ln \left( \frac{z - 2}{z + 2} \right) = \ln x + C$$ $$\ln \left( \frac{z - 2}{z + 2} \right) = \ln x^4 + C$$ $$\frac{z - 2}{z + 2} = C x^4$$ $$z - 2 = C x^4(z + 2)$$ $$z(1 - C x^4) = 2C x^4 + 2$$ $$z = \frac{2C x^4 + 2}{1 - C x^4}$$

Get back to substitute

$$x y = \frac {2 C x ^ {4} + 2}{1 - C x ^ {4}}$$ $$y = \frac {C x ^ {4} + 2}{x (1 - C x ^ {4})}$$

Answer:

$$y = \frac {C x ^ {4} + 2}{x (1 - C x ^ {4})}$$

2.

Solution

$$y = x y ^ {\prime} + 1 - \ln y ^ {\prime}$$

This is the Clero equation of the form $y = x y' + f(y')$ . Parametrizize

$$\left[ \begin{array}{c} y ^ {\prime} = p \\ x = x \\ y = x p + 1 - \ln p \\ d y = y ^ {\prime} d x \end{array} \right.$$ $$x d p + p d x - \frac {d p}{p} = p d x$$ $$d p \left(x - \frac {1}{p}\right) = 0$$ $$d p = 0 \quad x - \frac {1}{p} = 0$$ $$p = C \quad p = \frac {1}{x}$$ $$y = C x + 1 - \ln C \quad y = 2 + \ln x$$

Answer:

$$\left[ \begin{array}{l} y = C x + 1 - \ln C \\ y = 2 + \ln x \end{array} \right.$$

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