Thread: Simple Initial Value Problem..

hi guys,
ok, so im reading my lecture notes again and here goes..

Solve the IVP dx/dt = 2x^2 with condition x(5) = -3

Solution: x(t) = -1/c + 2t
At t=5, we require x(5) = -3 (not equal to 0). We can use the solution x(t) = -1/c+2t, and determine c from the data given, i.e. -3 = -1/c+2t
so c = -29/3

now this is where i dont understand! (how did they get 6t ?? )

and hence the solution on the IVP is given by x(t) = 3/(29 - 6t)


thanks guys!!

Originally Posted by matlabnoob

hi guys,
ok, so im reading my lecture notes again and here goes..

Solve the IVP dx/dt = 2x^2 with condition x(5) = -3

Solution: x(t) = -1/c + 2t
At t=5, we require x(5) = -3 (not equal to 0). We can use the solution x(t) = -1/c+2t, and determine c from the data given, i.e. -3 = -1/c+2t
so c = -29/3

now this is where i dont understand! (how did they get 6t ?? )

and hence the solution on the IVP is given by x(t) = 3/(29 - 6t)


thanks guys!!

I think there's a problem. . When you spearate

so or , not what you have.

aah of course ,how did i miss that?

so its -1/c'+2t..

im trying to get 6t somehow as the result does... is it possible to get 6t? =S because what im doing is plugging in c = -29/3 into x(t) and t=5 into x(t)... but im not getting any 6ts..

Originally Posted by Danny

I think there's a problem. . When you spearate

so or , not what you have.

Now use the IC so giving

Therefore .

Originally Posted by matlabnoob

hi guys,
ok, so im reading my lecture notes again and here goes..

Solve the IVP dx/dt = 2x^2 with condition x(5) = -3

Solution: x(t) = -1/c + 2t
At t=5, we require x(5) = -3 (not equal to 0). We can use the solution x(t) = -1/c+2t, and determine c from the data given, i.e. -3 = -1/c+2t
so c = -29/3

now this is where i dont understand! (how did they get 6t ?? )

and hence the solution on the IVP is given by x(t) = 3/(29 - 6t)


thanks guys!!

Alternative method:















.

, where .


Now by using the fact that







.


Therefore



.