Question

Solve the following with the help of Banoulli equation?

dy/dx + y/2x = x/y^3

Answer 1

Answer on Question #52703- Math - Differential Calculus | Equations

Solve the following by Bernoulli equation

$$\frac{dy}{dx} + \frac{y}{2x} = \frac{x}{y^3}$$ $$\frac{dy}{dx} + \frac{y}{2x} = \frac{x}{y^3}$$

Solution

Let $z = y^{1 - (-3)} = y^4$, then

$$\frac{dz}{dx} = \frac{4y^3}{dy}$$ $$\frac{dy}{dx} = \frac{1}{4y^3} \frac{dz}{dx}$$ $$\frac{1}{4y^3} \frac{dz}{dx} + \frac{y}{2x} = \frac{x}{y^3}$$ $$\frac{dz}{dx} + \frac{4y^4}{2x} = 4x$$ $$\frac{dz}{dx} + \frac{2x}{x} = 4x$$

This is a linear first order ordinary differential equation in the dependent variable $x$. We write the equation in the form:

$$\frac{dz}{dx} + P(x)z = Q(x)$$ $$\frac{dz}{dx} + \frac{2x}{x}z = 4x$$

So $P(x) = \frac{2}{x}, Q(x) = 4x$

The integrating factor is

$$\mu(x) = e^{\int P(x)dx}$$

that is,

$$\mu(x) = e^{\int \frac{2}{x} dx}$$ $$\int \frac{2x}{x} dx = 2\ln x$$ $$\mu(x) = e^{2\ln x} = x^2$$

Consider a differential equation

$$\frac{dz}{dx} + \frac{2x}{x}z = 4x$$

multiplying by $x^{2}$ gives

$$x^{2} \frac{dz}{dx} + 2xz = 4x^{3}$$

so, $(x^{2}z)^{'} = 4x^{3}$

integrating both sides with respect to $x$ gives

$$x^{2}z = x^{4} + C,$$

whence

$$z = x^{2} + \frac{C}{x^{2}},$$

where $C$ is an arbitrary real constant.

Recalling the substitution $z = y^4$, obtain that

$$y^{4} = x^{2} + \frac{C}{x^{2}},$$

where $C$ is an arbitrary real constant.

The solution to a linear first order differential equation is then

$$z(x) = \frac{\int \mu(x) Q(x) dx + C}{\mu(x)}$$ $$z(x) = \frac{\int 4x^{3} dx + C}{x^{2}} = \frac{x^{4} + C}{x^{2}} = x^{2} + \frac{C}{x^{2}},$$

where $C$ is an arbitrary real constant.

Answer:

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

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