Math / Calculus

Question1. Solve this DE using Exact Techrique $\frac{d y}{d x}=\frac{y-y^{2}}{2 x y+1}, \quad y>1$

Studdy Solution

STEP 1

Assumptions
1. The differential equation is given by: $\frac{d y}{d x} = \frac{y - y^2}{2xy + 1}$
2. We need to solve this differential equation using the Exact Technique.
3. The condition given is $y > 1$ .

STEP 2

To solve the differential equation using the Exact Technique, we first need to express it in the form: $M(x, y) + N(x, y) \frac{dy}{dx} = 0$
Rearrange the given differential equation: $(2xy + 1) \, dy = (y - y^2) \, dx$

STEP 3

Identify $M(x, y)$ and $N(x, y)$ from the rearranged equation: $M(x, y) = -(y - y^2)$ $N(x, y) = 2xy + 1$

STEP 4

Check if the differential equation is exact by verifying if: $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$
Calculate $\frac{\partial M}{\partial y}$ : $\frac{\partial M}{\partial y} = -1 + 2y$

STEP 5

Calculate $\frac{\partial N}{\partial x}$ : $\frac{\partial N}{\partial x} = 2y$

STEP 6

Compare $\frac{\partial M}{\partial y}$ and $\frac{\partial N}{\partial x}$ : $\frac{\partial M}{\partial y} = -1 + 2y$ $\frac{\partial N}{\partial x} = 2y$
Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$ , the equation is not exact.

STEP 7

Since the equation is not exact, we need to find an integrating factor to make it exact. An integrating factor can often be a function of $x$ or $y$ alone.

STEP 8

Assume an integrating factor $\mu(y)$ that depends only on $y$ . The condition for $\mu(y)$ is: $\frac{\partial}{\partial y}(\mu(y) M) = \frac{\partial}{\partial x}(\mu(y) N)$

STEP 9

Substitute $M$ and $N$ into the condition: $\frac{\partial}{\partial y}(\mu(y)(-y + y^2)) = \frac{\partial}{\partial x}(\mu(y)(2xy + 1))$

STEP 10

Calculate the left side: $\frac{\partial}{\partial y}(\mu(y)(-y + y^2)) = \mu(y)(-1 + 2y) + \mu'(y)(-y + y^2)$

STEP 11

Calculate the right side: $\frac{\partial}{\partial x}(\mu(y)(2xy + 1)) = \mu(y)(2y)$

STEP 12

Equate the two expressions: $\mu(y)(-1 + 2y) + \mu'(y)(-y + y^2) = \mu(y)(2y)$

STEP 13

Simplify and solve for $\mu(y)$ : $\mu(y)(-1 + 2y) + \mu'(y)(-y + y^2) = \mu(y)(2y)$ $\mu'(y)(-y + y^2) = \mu(y)$

STEP 14

Separate variables and integrate: $\frac{\mu'(y)}{\mu(y)} = \frac{1}{-y + y^2}$ $\ln|\mu(y)| = \int \frac{1}{-y + y^2} \, dy$

STEP 15

Simplify the integral: $\int \frac{1}{-y + y^2} \, dy = \int \frac{1}{y(1-y)} \, dy$

STEP 16

Use partial fraction decomposition: $\frac{1}{y(1-y)} = \frac{A}{y} + \frac{B}{1-y}$ Solving gives $A = 1$ and $B = 1$ .

STEP 17

Integrate the partial fractions: $\int \left( \frac{1}{y} + \frac{1}{1-y} \right) \, dy = \ln|y| - \ln|1-y|$

STEP 18

Substitute back to find $\mu(y)$ : $\ln|\mu(y)| = \ln|y| - \ln|1-y|$ $\mu(y) = \frac{y}{1-y}$

STEP 19

Multiply the original differential equation by the integrating factor $\mu(y) = \frac{y}{1-y}$ to make it exact.

STEP 20

The modified equation becomes: $\frac{y}{1-y}(-y + y^2) \, dx + \frac{y}{1-y}(2xy + 1) \, dy = 0$

STEP 21

Simplify the equation: $(-y^2 + y^3) \, dx + \left(\frac{2xy^2}{1-y} + \frac{y}{1-y}\right) \, dy = 0$

STEP 22

Now, the equation is exact. Integrate $M$ with respect to $x$ : $\int (-y^2 + y^3) \, dx = -xy^2 + xy^3 + C(y)$

STEP 23

Differentiate the result with respect to $y$ and equate to $N$ : $\frac{\partial}{\partial y}(-xy^2 + xy^3 + C(y)) = -2xy + 3xy^2 + C'(y)$

STEP 24

Equate to $N$ : $-2xy + 3xy^2 + C'(y) = \frac{2xy^2}{1-y} + \frac{y}{1-y}$

STEP 25

Solve for $C'(y)$ : $C'(y) = \frac{2xy^2}{1-y} + \frac{y}{1-y} + 2xy - 3xy^2$

STEP 26

Integrate $C'(y)$ with respect to $y$ to find $C(y)$ .

STEP 27

Substitute $C(y)$ back into the expression for the solution.

STEP 28

The solution to the differential equation is given by: $-xy^2 + xy^3 + C(y) = \text{constant}$

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