- #1
braindead101
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Find the solution to the initial value problem.
2ty^3 + 3t^2y^2 dy/dt = 0 , y(1) = 1
I found out whether the equation was exact and it was, and i continued as follows. NOTE: the answer is y(t) = t^(-4/3) i just don't know how to get it. or maybe my solution is just in another form.
im going to use f instead of that greek symbol i do not how to type out here.
M(t,y) = 2ty^3
N(t,y) = 3t^2y^2
don't know why, just copied from textbook.
M(t,y) = df(t,y)/dt = 2ty^3
N(t,y) = df(t,y)/dy = 3t^2y^2
f(t,y) = integ(M(t,y)dt) + h(y)
f(t,y) = integ(2ty^3 dt) + h(y)
f(t,y) = 2y^3(1/2t^2) + h(y)
f(t,y) = y^3t^2 + h(y)
df(t,y)/dy = t^2(3y^2) + dh(y)/dy
df(t,y)/dy = 3t^2y^2 + dh(y)/dy
3t^2y^2 = 3t^2y^2 + dh(y)/dy
dh(y)/dy = 1
h(y) = integ(1 dy) + c
h(y) = y + c
f(t,y) = y^3t^2 + y + c
y^3t^2 + y = C
sub y(1) = 1
(1)^3(1)^2 + 1 = C
C= 2
.'. y(t)^3t^2 + y(t) = 2
and i can't isolate y(t) so i left that as final solution. any thoughts on whether this is right or wrong.
2ty^3 + 3t^2y^2 dy/dt = 0 , y(1) = 1
I found out whether the equation was exact and it was, and i continued as follows. NOTE: the answer is y(t) = t^(-4/3) i just don't know how to get it. or maybe my solution is just in another form.
im going to use f instead of that greek symbol i do not how to type out here.
M(t,y) = 2ty^3
N(t,y) = 3t^2y^2
don't know why, just copied from textbook.
M(t,y) = df(t,y)/dt = 2ty^3
N(t,y) = df(t,y)/dy = 3t^2y^2
f(t,y) = integ(M(t,y)dt) + h(y)
f(t,y) = integ(2ty^3 dt) + h(y)
f(t,y) = 2y^3(1/2t^2) + h(y)
f(t,y) = y^3t^2 + h(y)
df(t,y)/dy = t^2(3y^2) + dh(y)/dy
df(t,y)/dy = 3t^2y^2 + dh(y)/dy
3t^2y^2 = 3t^2y^2 + dh(y)/dy
dh(y)/dy = 1
h(y) = integ(1 dy) + c
h(y) = y + c
f(t,y) = y^3t^2 + y + c
y^3t^2 + y = C
sub y(1) = 1
(1)^3(1)^2 + 1 = C
C= 2
.'. y(t)^3t^2 + y(t) = 2
and i can't isolate y(t) so i left that as final solution. any thoughts on whether this is right or wrong.