Question

Let x ln (y)+k y²=k (for .k=39) and y'=-ln (y)/x/y+m y then m is equal to

Answer 1

The value of m in the differential equation y' = -\ln(y)/(xy) + my is calculated by isolating \ln(y) from the first equation and plugging it into the second equation, yielding m = -39(1 - y^2)/(x^2y^2), which depends on the values of x and y.

Step-by-step explanation:

The given equations are x \ln(y) + ky^2 = k where k = 39, and y' = -\frac{\ln(y)}{xy} + my. We are asked to determine the value of m that would make the second equation true considering the first one.

To find m, rewrite the first equation by isolating \ln(y) on one side:

\ln(y) = \frac{k - ky^2}{x} = \frac{k(1 - y^2)}{x}.

Now, plug this into the derivative given in the second equation:

y' = -\frac{k(1 - y^2)}{x^2y} + my.

According to the second equation, the term that multiplies y should be m. Therefore, by comparing the terms, we conclude that:

m = -\frac{k(1 - y^2)}{x^2y^2}

Plugging in k = 39:

m = -\frac{39(1 - y^2)}{x^2y^2}

So, the value of m depends on the values of x and y decided by the context of the problem.