Question

E^((2x^2)+(2y^2))=xe^(29y)-y^2e^(58x/5) ; (5,2)

Answer 1

To solve the equation e^((2x^2)+(2y^2))=xe^(29y)-y^2e^(58x/5), isolate the variables x and y. Take the natural logarithm of both sides, collect like terms, and rearrange the equation. Solve the quadratic equation to find the values of x and y that satisfy the equation.

Step-by-step explanation:

To solve the equation e^((2x^2)+(2y^2))=xe^(29y)-y^2e^(58x/5), we need to isolate the variables x and y. Let's break it down step by step:

1. Start by taking the natural logarithm of both sides of the equation to remove the exponential terms.
2. You will end up with (2x^2) + (2y^2) = ln(x) + (29y) - ln(y^2) - (58x/5).
3. Next, collect all the like terms and rearrange the equation to get 2x^2 + 58x/5 + ln(x) - (2y^2) - (29y) + ln(y^2) = 0.
4. As the equation is quadratic in nature, you can now use various methods to solve it, such as factoring, completing the square, or using the quadratic formula.

From here, you can work towards finding the values of x and y that satisfy the equation. Plug in the given values (5,2) for x and y into the equation and see if it holds true.