Math Problem Statement

(c) x(e

ସ௬ − 1) ௗ௬

ௗ௫

  • (x

ଶ − 1)e

ଶ௬ = 0, x > 0 solve using separation of variables

Solution

We are given the differential equation:

xeydydx+(x21)e2y=0,x>0x e^{y} \frac{dy}{dx} + (x^2 - 1)e^{2y} = 0, \quad x > 0

We will solve this using the method of separation of variables.

Step 1: Rewriting the equation

We start by rewriting the equation:

xeydydx=(x21)e2yx e^{y} \frac{dy}{dx} = -(x^2 - 1) e^{2y}

Now, divide both sides by eye^y:

xdydx=(x21)eyx \frac{dy}{dx} = -(x^2 - 1) e^y

Step 2: Separate the variables

We want to separate the variables xx and yy. So, divide both sides by (x21)-(x^2 - 1) and multiply by dxdx:

dyey=(x21)xdx\frac{dy}{e^y} = - \frac{(x^2 - 1)}{x} dx

Step 3: Integrate both sides

Now we can integrate both sides. The left side requires integrating with respect to yy, and the right side with respect to xx.

For the left side:

dyey=(x21)xdx\int \frac{dy}{e^y} = \int - \frac{(x^2 - 1)}{x} dx

The left-hand side simplifies to:

eydy=ey\int e^{-y} dy = -e^{-y}

For the right-hand side, we simplify the integrand:

(x21)xdx=x+1xdx\int - \frac{(x^2 - 1)}{x} dx = \int -x + \frac{1}{x} dx

Now integrate:

xdx+1xdx=x22+lnx- \int x dx + \int \frac{1}{x} dx = - \frac{x^2}{2} + \ln |x|

Step 4: Combine the results

We now combine both results from the integrals:

ey=x22+lnx+C-e^{-y} = - \frac{x^2}{2} + \ln |x| + C

Where CC is the constant of integration. Simplifying, we get:

ey=x22lnxCe^{-y} = \frac{x^2}{2} - \ln |x| - C

Step 5: Solve for yy

Now, solve for yy by taking the natural logarithm of both sides:

y=ln(x22lnxC)y = - \ln \left( \frac{x^2}{2} - \ln |x| - C \right)

Thus, the solution to the differential equation is:

y(x)=ln(x22lnxC)y(x) = - \ln \left( \frac{x^2}{2} - \ln |x| - C \right)

Summary

Using separation of variables, we have solved the differential equation and obtained the general solution for yy in terms of xx.

Would you like more details on any of the steps?

Here are some related questions to deepen your understanding:

  1. How can separation of variables be applied to other types of differential equations?
  2. What is the importance of the constant of integration in solving differential equations?
  3. How do you handle boundary conditions or initial conditions when solving differential equations?
  4. Why is it necessary to check the domain of the solution after solving?
  5. Can the method of separation of variables be extended to partial differential equations?

Tip: Always check the consistency of units or dimensions when solving physical problems involving differential equations.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Separation of Variables
Integration Techniques

Formulas

\frac{dy}{dx} = -(x^2 - 1) e^y / x
\int \frac{dy}{e^y} = \int - \frac{(x^2 - 1)}{x} dx
e^{-y} = \frac{x^2}{2} - \ln |x| - C

Theorems

Separation of Variables Method
Integration by Substitution

Suitable Grade Level

College Level