$$x = c_1\cos (t) + c_2\sin (t) $$ is a two-parameter family of solutions of the second-order DE $$x'' + x = 0. $$

Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions:

$$x\left(\frac{\pi}{6}\right) = \frac{1}{2} $$


$$x'\left(\frac{\pi}{6}\right) = 0 $$

I found that:

$$x'(t) = - c_1\sin (t) + c_2\cos (t) $$

and then I solved for $c_2$

$$c_2 = 1 - c_1\sqrt 3 $$

and plugged $c_2$ into $$x'\left(\frac{\pi}{6}\right) = 0 $$


$$x'\left(\frac{\pi}{6}\right) = - c_1\sin (t) + (1 - c_1\sqrt 3 )\cos (t) = 0 $$

and $$ = \frac{1}{2}c_1 + 1 - c_1\frac{{\sqrt 3 }}{2} = 0 $$

and finally got as far as

$$c_1 = \frac{{(\frac{1}{2} - \frac{{\sqrt 3 }}{2})}}{{\frac{1}{2} - \frac{{\sqrt 3 }}{2}}} = \frac{{ - 1}}{{\frac{1}{2} - \frac{{\sqrt 3 }}{2}}} = $$

However the book's solutions guide came up with

$$c_1 = \frac{{\sqrt 3 }}{4},c_2 = \frac{1}{4} $$


$$\frac{{\sqrt 3 }}{2}c_1 + \frac{1}{2}c_2 = \frac{1}{2} $$


$$ - \frac{1}{2}c_1 + \frac{{\sqrt 3 }}{2}c_2 = 0 $$

I just can't follow their logic, So any help would be appreciated.


From the equations $$ x'\left(\frac{\pi }{6}\right) = 0 \tag1 $$ and $$ x'(t) = - c_1\sin (t) + c_2\cos (t) \tag2 $$ you can set $t = \frac\pi6$ (which I assume is what you did) so that $x'(t)=0,$ $\sin (t)=\frac12,$ and $\cos (t) = \frac{\sqrt3}2,$ and plugging these values into Equation $(2)$ you would get $$ 0 = -c_1 \cdot \frac12 + c_2 \cdot \frac{\sqrt3}2. \tag3 $$ Solving Equation $(3)$ for $c_2$ in terms of $c_1,$ your first step might be to bring $c_1$ to the left hand side, so that $$ c_1 \cdot \frac12 = c_2 \cdot \frac{\sqrt3}2. $$ and then multiply by $\frac2{\sqrt3}$ on both sides to get $$ \frac1{\sqrt3} c_1 = c_2. \tag4 $$

Once you have $c_2$ in terms of $c_1$ (or vice versa) you still have to use the fact that $x\left(\frac{\pi }{6}\right) = \frac12,$ of course.

Actually it was unclear what you meant by "then I solved for $c_2$". I assumed you meant you solved the immediately preceding equation with $x'(t),$ but if you meant that you went all the way back to the first equation you wrote and solved that at $t = \frac\pi6,$ you would get $$ \frac12 = c_1 \frac{\sqrt3}2 \cdot + c_2 \cdot \frac12. \tag5 $$ And now if you solve for $c_2$ you do indeed get $c_2 = 1 - c_1 \sqrt3.$ But in the subsequent steps you should have found that when $t = \frac\pi6,$ \begin{align} 0 &= - c_1\sin(t) + (1 - c_1\sqrt3)\cos(t) \\ &= -c_1\cdot\frac12 + (1 - c_1\sqrt3)\cdot \frac{\sqrt3}2 \\ &= -\frac12 c_1 + \frac{\sqrt3}2 - \frac32 c_1 \\ &= \frac{\sqrt3}2 - 2 c_1, \end{align} and that is how we end up with $c_1 = \frac{\sqrt3}4.$

Now for what the book's author(s) apparently did: they derived Equation $(5)$ using the same method I described above, except that they wrote the equation in the form $$\frac{\sqrt3}2 c_1 + \frac12 c_2 = \frac12. \tag6$$ Then they derived Equation $(3)$ just as I did above, except they wrote it $$ - \frac12 c_1 + \frac{\sqrt3}2 c_2 = 0. \tag7$$ Now Equations $(6)$ and $(7)$ are two linear equations in two unknowns and there are standard methods (which you should study if you don't know them already) for solving such a set of equations. For example, multiply both sides of Equation $(7)$ by $\sqrt3$ so that when you add the left side of Equation $(7)$ to the left side of Equation $(6)$ the terms in $c_1$ cancel and you're left with $2c_2$; while adding the right-hand sides of these equations gives you $\frac12,$ so you conclude that $$2c_2 = \frac12$$ and therefore $c_2 = \frac14.$ Plug that into either equation and solve for $c_1.$ Of course if you notice that Equation $(7)$ has no constant term, you may see that it immediately gives you $c_1 = (\sqrt3)c_2,$ which you can plug into Equation $(6)$ to solve for $c_2.$

A word of advice: lay out your steps much more carefully and methodically and be very clear about what you're doing and why at each step. Whoever grades your papers will appreciate it.

  • $\begingroup$ Thankyou for the insight and help with this problem! $\endgroup$ – HappyHiggs Jun 11 at 2:34

Given that $x=c_1\cos t+c_2\sin t$ which are the family of solutions.

Finding the first derivative of the solution

$$x^{\prime}=-c_1\sin t+c_2\cos t$$

Now Substitute the initial condition of $x\left(\dfrac{\pi}{6}\right)=\dfrac12$

$$\dfrac12=c_1\cos\left(\dfrac{\pi}{6}\right)+c_2\sin\left(\dfrac{\pi}{6}\right)$$ $$\dfrac{\sqrt3}{2}c_1+\dfrac12c_2=\dfrac12\tag1$$

Now Substitute the initial condition of $x^\prime\left(\dfrac{\pi}{6}\right)=0$ $$0=-c_1\sin\left(\dfrac{\pi}{6}\right)+c_2\cos\left(\dfrac{\pi}{6}\right)$$ $$-\dfrac12c_1+\dfrac{\sqrt3}{2}c_2=0\tag2$$

Solving $(1)$ and $(2)$ we find the following values for $c_1$ and $c_2$ $$c_1=\dfrac{\sqrt3}{4},\ \ c_2=\dfrac12$$

Therefore the solution is $x=\dfrac{\sqrt3}{4}\cos t+\dfrac12\sin t$

  • $\begingroup$ Thankyou very much $\endgroup$ – HappyHiggs Jun 11 at 2:33
  • $\begingroup$ @HappyHiggs If you find the answer helpful then accept the answer $\checkmark$ :) $\endgroup$ – Key Flex Jun 11 at 15:47

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