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I read this question in my math textbook. But even its solution I can't even understand. This is in the introduction to differential equation. So I have't yet learnt any formal way to solve DE.

$$\frac{d^2y}{dx^2}=-ky$$ Step 1 (Multiply each side by $2\frac{dy}{dx}$): $$2\frac{d^2y}{dx^2}\frac{dy}{dx}=-2ky\frac{dy}{dx}$$ Step 2 (Integrate both sides) And somehow get: $$(\frac{dy}{dx})^2=-ky^2+C$$ Step 3 I don't know. Are you supposed to take the square root?

I am confused as how the author get from Step 1 to Step 2. I know for the right hand side, he integrate with respect to $y$. But what happen the left hand side? And how do you solve the rest of it?

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    Do you know the chain rule? – 2017-01-28
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    In which text book was this written? – 2017-01-28
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    Morris Kline An intuitve approach to Calc – 2017-01-28
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    I do know the chain rule. The basic version. How things cancel out. – 2017-01-28

3 Answers 3

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From step 2 to step 3 he integrated both sides with respect to $x$, not $y$. You can check it by differentiating step 3 with respect to $x$. On the left, you have $(\frac {dy}{dx})^2,$, which differentiates by the chain rule to $2(\frac {dy}{dx})(\frac {dy}{dx})'=2(\frac {dy}{dx})(\frac {d^2y}{dx^2})$ On the right you differentiate $-ky^2+C$ to $-2kyy'$

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    I agree with you about the left side. But the right is take derivative of with respect to y not x. – 2017-01-28
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    No, if you took the derivative with respect to $y$ you would just get $-2ky$. The $y'$ factor comes from the chain rule when you differentiate with respect to $x$ – 2017-01-28
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    Oh yeah...Thanks. – 2017-01-29
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$$y'' + ky = 0$$

The characteristic equation reads

$$\lambda^2 + k = 0$$

Hence

$$\lambda_{1, 2} = \pm i\sqrt{k}$$

hence the solution follows

$$y(x) = C_1 e^{-i\sqrt{k}x} + C_2 e^{ i \sqrt{k}x}$$

Or if you prefer

$$y(x) = C_1 \cos\sqrt{k}x + C_2 \sin\sqrt{k}x$$

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    I havent learn about characteristic equation? Is this the solution to $(\frac{dy}{dx})^2=-ky^2+C$? – 2017-01-28
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    It's the solution of the equation you wrote at the beginning! – 2017-01-28
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    the same thing. – 2017-01-28
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Here is another approach which gives the answer directly. Your approach gives the answer in terms of $x$ as a function of $y$.

The problem is simpler if $k$ is positive. Let $k = a^{2} > 0$ and then we put $t = ax$ so that $$\frac{dy}{dx} = a\frac{dy}{dt},\frac{d^{2}y}{dx^{2}} = a^{2}\frac{d^{2}y}{dt^{2}} = k\frac{d^{2}y}{dt^{2}}$$ so that the equation is transformed into $$\frac{d^{2}y}{dt^{2}} + y = 0\tag{1}$$ Let the initial conditions be $y(0) = A, y'(0) = B$ (here prime denotes derivative with respect to $t$). Then we show that $y = f(t) = A\cos t + B\sin t$ for all real $t$.

To do so first we note that the function $h(t) = f(t) - A\cos t - B\sin t$ also satisfies $h''(t) + h(t) = 0$ and $h(0) = h'(0) = 0$. Consider the function $$F(t) = \{h(t)\}^{2} + \{h'(t)\}^{2}$$ Clearly we have $$F'(t) = 2h'(t)\{h(t) + h''(t)\} = 0$$ and hence $F$ is constant so that $F(t) = F(0) = 0$. Thus we can see that $$\{h(t)\}^{2} + \{h'(t)\}^{2} = 0$$ for all real $t$. Since $h, h'$ are real it follows that $h(t) = 0$ for all real $t$ and hence $f(t) = A\cos t + B\sin t$.

Transforming this to variable $x$ we can see that the solution to the original equation is given by $$y = y(0)\cos \sqrt{k}x + \frac{y'(0)}{\sqrt{k}}\sin\sqrt{k}x$$

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    Why the downvote? – 2017-01-29