Can anyone help me solve double derivative equation? $d^2x=kd^2y$

Question

My approach:

$$d^2x=k*d^2y$$ $$\int d^2x=\int k*d^2y$$ $$dx+c_1=k*dy+c_2$$ $$dx=k*dy+c_2-c_1$$ $$\int dx=\int (k*dy+c_2-c_1)$$ $$x=ky+(c_2-c_1)+c_3$$ $$x=ky+c$$

I'm learning maths and algebra, online, without a tutor. Sorry.

Answer 1

The notation for second derivative is $\frac{d^2y}{dx^2}$ NOT $\frac{d^2y}{d^2x}$. See: https://en.wikipedia.org/wiki/Second_derivative

You can write $$\frac{d^2y}{dx^2}=\frac{d(\frac{dy}{dx})}{dx}=\frac{1}{k}\tag{1}$$ to solve the equation $y''=\frac{1}{k}$ which is the equation $\frac{d^2y}{dx^2}=\frac{1}{k}$ but NOT the equation $d^2x=kd^2y$.

From (1), $d(\frac{dy}{dx})=\frac{1}{k}dx$. By integrating you have $\frac{dy}{dx}=\frac{1}{k}x+c$. Then, $dy=(\frac{1}{k}x+c)dx$ and by integrtaing again $y=\frac{1}{2k}x^2+cx+d$ where $c$ and $d$ are constants.

Solution for your equation (My opinion): As suggested by J.G., we can write your equation as $d^2(x-ky)=0$. Then $d(x-ky)=c$, a constant. Here, the left hand side, $d(x-ky)$ is infinite small. So, $c$ must be zero. We get $d(x-ky)=0$. Hence, $x-ky=c$ where c is a constant.