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This problem is in Penney's Elementary Differential Equations, listed as a reducible, 2nd-order DE. The chapter has taught two techniques to be used, which are for when either $x$ or $y(x)$ is missing. It didn't show how to solve when $y'(x)$ is missing.

I checked with WolframAlpha, and it suggests starting by assuming that $y$ is proportional to $e^{\lambda x}$.

If I didn't know to assume this, how could I solve this otherwise?

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    $\begingroup$ There is a known method for differential equations in which the coefficients are constant. Edit: Check the section 2.3. Homogeneous equations with constant coefficients in your book. $\endgroup$ – Git Gud Jan 30 '14 at 22:59
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    $\begingroup$ I agree that that assumption seems mysterious. My way of solving this would be to remember that $\sin''=-\sin$ and $\cos''=-\cos$, and then recall theoretical reasons why the solution space should be $2$-dimensional. The factor of $4$ is easily adjusted for: $(d/dx)\sin(4x)=-4\sin(4x)$ etc. $\endgroup$ – Michael Hardy Jan 30 '14 at 23:08
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$y''+4y=0$ is a second order differential equation. First, change the equation to

$r^2+4 = 0$

This equation will will have complex conjugate roots, so the final answer would be in the form of $e^{\alpha x}(c_1\sin(\beta x) + c_2 \cos (\beta x))$ where $\alpha$ equals the real part of the complex roots and $\beta$ equals the imaginary part of (one of) the complex roots.

We need to use the quadratic formula \begin{array}{*{20}c} {x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0} \\ \end{array}

In this equation $a =1$, $b=0$, and $c =4$

\begin{array}{*{20}c} {x = \frac{{ -0 \pm \sqrt {0^2 - 4(1)(4)} }}{{2}}} \\ \end{array} \begin{array}{*{20}c} {x = \frac{{ -0 \pm \sqrt {-16} }}{{2}}} \\ \end{array} \begin{array}{*{20}c} {x = \frac{{\pm \sqrt {-16} }}{{2}}} \\ \end{array} Since we can't have negative signs in the square root, we have an imaginary number $i$.

\begin{array}{*{20}c} {x = \frac{{\pm \sqrt {16}i }}{{2}}} \\ \end{array} Take the square root \begin{array}{*{20}c} {x = \frac{{\pm 4i }}{{2}}} \\ \end{array} \begin{array}{*{20}c} {x = \pm 2i} \\ \end{array} Now let's bring this equation back.

$y = e^{\alpha x}(c_1 \sin(\beta x) + c_2 \cos (\beta x))$

$\alpha = 0$ and $\beta = 2$

So your answer would be

$y = e^{0x}(c_1\sin(2x) + c_2\cos (2x))$

Since $e^{0x}$ is $1$ because $0$ multiplied by $x$ is just $0$ and $e^{0}$ is $1$.

The final answer is $y = (c_1\sin(2x) + c_2\cos (2x))$.

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    $\begingroup$ Why use the quadratic formula? $r^2 +4 = 0$ implies $r^2 = -4$ which implies $r=\pm 2i$... $\endgroup$ – fretty Jan 31 '14 at 11:03
  • $\begingroup$ You could do that too, but most of the time I used the quadratic formula....actually nobody brought that suggestion up at all in class last semester. My professor used the quadratic formula. $\endgroup$ – usukidoll Feb 1 '14 at 3:19
  • $\begingroup$ Just seems like a bit of a sledgehammer way to do it rather than just rearranging and square rooting. Of course in general a 2nd order linear ODE will not always give a quadratic of this form so usually you must use the QF or similar methods but here the quadratic was an easy one. $\endgroup$ – fretty Feb 1 '14 at 8:13
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Use a "test solution": $y=e^{\lambda x}$. Then the characteristic equation is $\lambda^2+4=0$. Can you proceed from here?

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    $\begingroup$ Did you read my question? $\endgroup$ – Korgan Rivera Jan 30 '14 at 23:40
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    $\begingroup$ We need to find the characteristic equation before we can go any further and since it does come out to $\lambda^2+4=0$, you can't factor, so we need to use the quadratic formula. Think of this equation $y''+6y'+6y = 0 $. $y''$ means $r^2$...$y'$ means $r$...and $y$ means well it has no meaning so you just write the number. $y''+6y'+6y = 0 $ becomes $r^2+6r+6 = 0$. Since the $y'$ isn't in your question, you can omit that part. $\endgroup$ – usukidoll Jan 31 '14 at 0:00
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    $\begingroup$ @KorganRivera: yes I did- but this is the standard method for doing the problem. $\endgroup$ – voldemort Jan 31 '14 at 0:42
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We can just factor it and solve two first order ordinary differential equations instead. Write $$y''+4y=y''+2iy'-2iy'+4y=(y'+2iy)'-2i(y'+2iy)=0.$$ Substitute $z=y'+2iy$ and get $z'-2iz=0\implies z(t)=A e^{2it}$. Substitute back: $$y'+2iy=Ae^{2it}\iff(e^{2it}y)'=Ae^{4it}$$ We then get the desired solution $$y=Ae^{2it}+Be^{-2it}.$$


Needless to say, the same technique works for any second order ordinary differential equation with constant coefficients. You can see a proof here, with special attention to the equal roots case.


Sorry, I just noted you are the OP of the linked question. If this answer is not useful, let me know.

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Put P=y'=dy/dx so y"=dP/dx=dP/dy * dy/dx =P dP/dx

y"+4y=0 <=> P dP/dy +4y = 0 <=> P dP = -4y dy <=> (P^2)/2 = -2(y^2) + C <=> P=(C-4y(^2))^(1/2)

P=dy/dx so dy/dx=(C-4y(^2))^(1/2)

dy / (C-4y(^2))^(1/2) = dx

put K^2=C so

(C-4y(^2))^(1/2) = K (1-4y^2/K^2)(^1/2)

and finally integration gives

Primitive of 1 / (K (1-4y^2/K^2)(^1/2)) = x

put u =2y/K so du/dy=2/K and x=1/K * primitive of (1/(1-u^2)^1/2)*K/2*du <=> x= 1/2 * primitive of (1/(1-u^2)^1/2)* du <=> x=1/2 arc sin u + L (L constant) <=> x - L = 1/2 arc sin u <=> 2x - 2L = arc sin u (put - 2L = F) <=> sin (2x + F) = u

u = 2y/K so <=> K/2 sin (2x + F) = y <=> K/2 (sin (2x) cos (F) + cox (2x) sin (F)) = y

Put A= K/2 cos (F) Put B= K/2 sin (F) General solution is y=A sin(2x) + B cos(2x)

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    $\begingroup$ please make use of MathJax (LaTeX renderer) which will help others to read your answer properly $\endgroup$ – user190080 Jul 14 '16 at 17:28

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