What can you conclude about the solution of y'=y^(2) just by looking? verify that y=-1/(x+c) are solution y'=y^(2). Find solution to the equation with the initial condition y(0)=1/3
I'm not sure what they want for the first part; I might say that the exponent of the solution is obviously -1 by the power rule... to check that the given formula is a solution just plug it into the original DE y'=y^(2) given that y=-1/(x+c) just plug in that y and check that the left side is the same as the right
sorry..could you explain more clear?
let y=-1/(x+c) what is y' ? what is y^2 ?
Ah ... sorry, m=y^2 ... try to plot y for different values of m |dw:1339183526251:dw|
nice piece of software http://www.math.psu.edu/cao/DFD/Dir.html you can even get solution here
to prove y=-1/(x+c) is solution of y' = y^2 plugin those values of y, i.e. \[ \frac d{dx} (-1/(x+c)) = \left ( \frac{-1}{x+c} \right )^2 \]
3) since you know y=-1/(x+c) is your solution, y(0) = -1/(0+c) = 3 find the value of c, and plugin the value of c, you have your particular solution
1) by looking i can say that your solution is gong to look like this |dw:1339184677172:dw|
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