Asked by Rpeep
Solve the initial-value problem for y as a function of x
(x^2+49)dy/dx=1
(x^2+49)dy/dx=1
Answers
Answered by
oobleck
dy = dx/(x^2+49)
recall that ∫ dx/(x^2+a^2) = 1/a arctan(x/a) + C
You have given no initial value, so that's as far as we can go for now.
recall that ∫ dx/(x^2+a^2) = 1/a arctan(x/a) + C
You have given no initial value, so that's as far as we can go for now.
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