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**Find the solution to the initial value problem.**

2ty^3 + 3t^2y^2 dy/dt = 0 , y(1) = 1

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.

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.