변수분리형 미분방정식

(a) If $y=5(e^x-3x^2-6x-6)$, then $\frac{\mathrm{d}y}{\mathrm{d}x} = 5(e^x-6x-6)$. Let's see whether this is equal to $y+15x^2$:

$$\begin{aligned} y+15x^2&=5(e^x-3x^2-6x-6)+15x^2\\ &=5(e^x-3x^2-6x-6+3x^2)\\ &=5(e^x-6x-6)\\ &=\frac{\mathrm{d}y}{\mathrm{d}x} \end{aligned}$$

So, $y=5(e^x-3x^2-6x-6)$ is indeed a solution to the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x}=y+15x^2$. (b) If $y=\dfrac{-2}{x^2+1}$, then $\frac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{4x}{(x+1)^2}$. Let's see whether this is equal to $xy^2$:

$$\begin{aligned} xy^2&=x\left(\frac{-2}{x^2+1}\right)^2\\ &=\frac{4x}{(x^2+1)^2}\\ &=\frac{\mathrm{d}y}{\mathrm{d}x} \end{aligned}$$

So, $y=\dfrac{-2}{x^2+1}$ is indeed a solution to the differential equation $\frac{\mathrm{d}y}{\mathrm{d}x}=xy^2$. (c) If $y=x^{3/2}+x$, then $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{3}{2}\sqrt{x}+1$.

$$\begin{aligned} \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2+\frac{\mathrm{d}y}{\mathrm{d}x}&=\left(\frac{3}{2}\sqrt{x}+1\right)^2+\frac{3}{2}\sqrt{x}+1\\ &=\frac{9}{4}x+\frac{9}{2}\sqrt{x}+2\\ &\neq \frac{\mathrm{d}y}{\mathrm{d}x} \end{aligned}$$

So, $y=x^{3/2}+x$ is not a solution to the differential equation $\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2+\frac{\mathrm{d}y}{\mathrm{d}x}=y$.