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ODE :- solve by reducing to exact form 2xy^2dx = e^x(dy-ydx)?
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ODE :- solve by reducing to exact form 2xy^2dx = e^x(dy-ydx)?
To solve the given ordinary differential equation (ODE) and reduce it to exact form, we will follow the steps below:

1. Understanding the ODE:
The given ODE is 2xy^2dx = e^x(dy - ydx).

2. Identifying the Form:
We observe that this ODE is not in exact form since the derivative of the coefficient of dx (2xy^2) with respect to y is not equal to the derivative of the coefficient of dy (e^x) with respect to x.

3. Rearranging the Equation:
Let's rearrange the equation by moving all the terms involving dy to the left side and all terms involving dx to the right side:
2xy^2 dx + ydx - e^x dy = 0

4. Identifying the Integrating Factor:
To convert this ODE into exact form, we need to find an integrating factor (IF). The IF is given by the exponential of the integral of the coefficient of dy with respect to y. In this case, the coefficient is -e^x, so the IF is e^(-x).

5. Multiplying the Equation by the Integrating Factor:
Multiply both sides of the equation by the IF (e^(-x)):
e^(-x) (2xy^2 dx + ydx - e^x dy) = 0

6. Rearranging the Equation:
Distribute e^(-x) to each term:
2xy^2 e^(-x) dx + ye^(-x) dx - e^(-x) e^x dy = 0

7. Simplifying the Equation:
Notice that e^(-x) e^x simplifies to 1, so we have:
2xy^2 e^(-x) dx + ye^(-x) dx - dy = 0

8. Identifying the Exact Form:
Now, the equation is in exact form, where M = 2xy^2 e^(-x) dx + ye^(-x) dx and N = -dy.

9. Finding the Total Differential:
To solve the ODE, we need to find a function f(x, y) such that df = M dx + N dy. We integrate M with respect to x and differentiate the result with respect to y to find f(x, y):
f(x, y) = ∫(2xy^2 e^(-x) + ye^(-x)) dx = x^2 y^2 e^(-x) + y^2 e^(-x) + g(y)

10. Solving for y:
To eliminate the constant of integration g(y), we differentiate f(x, y) with respect to y and set it equal to the given N:
∂f/∂y = 2xy^2 e^(-x) + 2y e^(-x) + ∂g/∂y = -1

11. Equating the Partial Derivatives:
Comparing the coefficients of y and the constant term on both sides, we have:
2xy^2 e^(-x) + 2y e^(-x) = 0
This equation can be simplified to:
2y(x + 1) = 0

12. Sol
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ODE :- solve by reducing to exact form 2xy^2dx = e^x(dy-ydx)?
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ODE :- solve by reducing to exact form 2xy^2dx = e^x(dy-ydx)? for Engineering Mathematics 2024 is part of Engineering Mathematics preparation. The Question and answers have been prepared according to the Engineering Mathematics exam syllabus. Information about ODE :- solve by reducing to exact form 2xy^2dx = e^x(dy-ydx)? covers all topics & solutions for Engineering Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ODE :- solve by reducing to exact form 2xy^2dx = e^x(dy-ydx)?.
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