Solution of differential with initial conditions

[kahwawashay1]

I think my book is giving me the wrong answer...The problem is to find solution of following:

r'(t) = t²$\hat{i}$ + 5t$\hat{j}$ + $\hat{k}$

The initial condition is:
r(1) = $\hat{j}$ + 2$\hat{k}$

My solution:

r(t) = < (1/3)t³ + c₁ , (5/2)t² + c₂ , t+c₃ >

r(1) = < 0 , 1 , 2 >
r(1) = < (1/3)+c₁ , (5/2)+c₂ , 1+c₃ >

Therefore:
< 0 , 1 , 2 > = < (1/3)+c₁ , (5/2)+c₂ , 1+c₃ >

Solving for the three c's yields:
c₁ = -(1/3)
c₂ = -1.5
c₃ = 1

And so the solution with the initial conditions is:
< (1/3)t³ - (1/3) , (5/2)t² -1.5 , t+1 >

My book gives the solution as:
< (1/3)t³ , (5/2)t² + 1 , t+2 >

Who is right?

 

[Mark44]

Your work looks fine to me. Possibly there is a typo in the book's solution, or maybe you are not working the same problem.

In the future, you can check these problems very easily. When you have your solution, check that
1) the initial condition is satisfied. For your problem, you're checking that r(1) = <0, 1, 2>, and
2) your solution satisfies the differential equation. Here, you're checking that r'(t) = <t^2, 5t, 1>.

 

[kahwawashay1]

Your work looks fine to me. Possibly there is a typo in the book's solution, or maybe you are not working the same problem.

In the future, you can check these problems very easily. When you have your solution, check that
1) the initial condition is satisfied. For your problem, you're checking that r(1) = <0, 1, 2>, and
2) your solution satisfies the differential equation. Here, you're checking that r'(t) = <t^2, 5t, 1>.

thanks!
yea i checked a thousand times if i am working the same problem as the book and i am.
i think the book meant to give the initial condition at r(0) not r(1)

this is like the third typo in this book so far..i can't believe i paid some 120$ for bunch of typos!