# Initial value Problem ODE not understanding solution

$$x = c_1\cos (t) + c_2\sin (t)$$ 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:

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

and

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

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

where

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

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

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

from

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

and

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

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.

• Thankyou for the insight and help with this problem! – 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 $$x=\dfrac{\sqrt3}{4}\cos t+\dfrac12\sin t$$

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