I need to find the maximal solution with
y(0)=1 and y(0)=-1
for the equation
y' = y^2 - 1
with t and y being real numbers.
By seperating so
y^2=1
and then integrating and isolating, I have found that
y = ±(3*t + 3*c)^(1/3)
where c is a constant, and to forfill the initial value problem needs to be 1/3, which gives me
y = ±(3*t + 1)^(1/3)
But I don't know if this is a maximal solution. How do I check if it is a maximal solution?
Initial value problem and maximal solution?
2016-06-05 10:11 am
回答 (2)
2016-06-05 10:57 am
✔ 最佳答案
y' = y^2 - 1 , well its a first order seperable ordinary differential equation,dy/dt = y^2 - 1 , now seperate the variables
(1/(y^2 - 1)) dy = dt , integrate both sides
∫ (1/(y^2 - 1) dy = ∫ dt , use partial fractions for the former,
1/(y^2 - 1) = A/(y-1) + B/(y+1) = (Ay + A + By - B) / (y^2 - 1) , so
A + B = 0
A - B = 1
A = 1/2 , B = -1/2 , so
(1/2) ∫ (1/(y-1) dy - (1/2) ∫ (1/(y+1)) dy = ∫ dt , so
(1/2)ln|y-1| - (1/2)ln|y+1| = t + c , by property of ln recall that, log(a/b) = log(a) - log(b) for all bases > 1,
(1/2)(ln|y-1| - ln|y+1|) = t + c,
(1/2)ln|(y-1)/(y+1)| = t + c ,
ln|(y-1)/(y+1)| = 2t + c, ofc its a new constant "c" i write c because -2*c is just a constant so i keep writing "c" for appearing new constants, exponentiate both sides,
(y-1)/(y+1) = c(e^(2t)) , i got here
y(t) would be
y(t) = (1 - c(e^(2t))) / (1 + c(e^(2t)))
收錄日期: 2021-04-21 18:54:34
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