Question

solve the given differential equation by using an appropriate substitution. the de is a bernoulli equation. dy dx = y(xy5 − 1)

Answer 1

For the positive case:

`\(y = (x + e^(x + C))^(-1/4)\)`

For the negative case:

`\(y = (x - e^(x + C))^(-1/4)\)`

These are the two solutions for the original differential equation.

To solve the given Bernoulli differential equation:

`\((dy)/(dx) = y(xy^5 - 1)\)`

We can use a substitution to transform it into a linear differential equation. Let's go through the steps:

Step 1: Identify the Bernoulli Equation Form

The given differential equation is in Bernoulli form because it is a first-order equation and can be written in the form:

`\((dy)/(dx) = P(x)y + Q(x)y^n\)`

In this case, P(x) = x and Q(x) = -1, and n = 5.

Step 2: Divide by
`y^n`

Divide both sides of the equation by
`\(y^5\):`

`\((1)/(y^5) (dy)/(dx) = x - (1)/(y^4)\)`

Step 3: Substitute a New Variable

Let
`\(z = y^(-4)\), then \((dz)/(dx) = -4y^(-5)(dy)/(dx)\).`Rewrite the equation in terms of z:

`\(-4(dz)/(dx) = x - z\)`

Step 4: Solve the Linear Differential Equation

The transformed equation is now a linear first-order differential equation. Solve it for z:

`\(-4(dz)/(dx) = x - z\)`

Rearrange the terms:

`\((dz)/(dx) = z - x\)`

Separate variables and integrate both sides:

`\((dz)/(z - x) = dx\)`

Integrate both sides:

`\(\int (dz)/(z - x) = \int dx\)`

This results in:

`\(\ln|z - x| = x + C\)`

where C is the constant of integration.

Step 5: Solve for z

Exponentiate both sides to solve for z:

`\(|z - x| = e^(x + C)\)`

Since the absolute value can be on the left side, we can express it as:

`\(z - x = \pm e^(x + C)\)`

Step 6: Remove Absolute Value

Remove the absolute value on the right side by considering both the positive and negative cases:

For the positive case:

`\(z - x = e^(x + C)\)`

For the negative case:

`\(z - x = -e^(x + C)\)`

Step 7: Express z in terms of y

Recall that
`\(z = y^(-4)\)`. Substitute this back in:

For the positive case:

`\(y^(-4) - x = e^(x + C)\)`

For the negative case:

`\(y^(-4) - x = -e^(x + C)\)`

Step 8: Solve for y

Now, we have two possible solutions for y:

For the positive case:

`\(y^(-4) = x + e^(x + C)\)`

Take the fourth root:

`\(y = (x + e^(x + C))^(-1/4)\)`

For the negative case:

`\(y^(-4) = x - e^(x + C)\)`

Take the fourth root:

`\(y = (x - e^(x + C))^(-1/4)\)`

These are the two solutions for the original differential equation.