How do you solve this differential equation: y' = x / y with initial value of y(1) = -2?

As stated, the equation is: y' = y / x



with initial value of y(1) = -2



The instructions for the problem from the text:

"..., find the exact solution of the initial value problem. ..."



The text is: Differential Equations, 2nd Edition

Authored: Polking, Boggess, and Arnold

Published: Pearson Prentice Hall



My process so far:



y' = y / x

dy / dx = y / x

(1 / y)*dy = (1 / x)*dx

鈭?1 / y)*dy = 鈭?1 / x)*dx

ln(y) = ln(x) + C (ignoring absolute values)

e^ln(y) = e^ln(x) + e^C

y = x + e^C



For init condition being y(1) = -2, this equation is unsolvable as it results in e^C = -3. So, obviously this is wrong.



The answer according to the book is:

y(x) = -2xHow do you solve this differential equation: y' = x / y with initial value of y(1) = -2?
I made the same mistake also as a student.



You almost have it, except at this step,



e^ln(y) = e^ln(x) + e^C



instead it should be



e^ln(y) = e^( ln(x) + c )



y = e^( ln(x) ) * e^c



y = x * C, where C is e^c



Now, use initial condition to solve for C



y(1) = -2



-2 = C,



Therefore, y(x) = -2x