differential equation $y^{\prime} = ky$

Question

I have the equation: \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}t} = -0.04\sqrt{y} \end{align*}

I had moved $\mathrm{d}t$ to the other side as well as $\sqrt{y}$ and took the integral on each side. I came to the equation \begin{align*} y = \left(\frac{-0.04t + c}{2}\right)^2 \end{align*}

But I am not sure this is correct and if c would just be y(0).

Answer 1

I am not actually sure which DE you want solved, so I'll solve both.

Number 1: $$f'(x)=kf(x)$$ $$\frac{f'(x)}{f(x)}=k$$ $$\int\frac{f'(x)}{f(x)}dx=k\int dx$$ Let $y=f(x)$. Therefore $dy=f'(x)dx$ $$\int\frac{dy}{y}=kx+c$$ $$\ln|y|=kx+c$$ $$f(x)=e^{kx+c}$$ Number 2: $$f'(x)=k\sqrt{f(x)}$$ $$\int\frac{f'(x)}{\sqrt{f(x)}}dx=kx+c$$ $$\int\frac{dy}{y^{1/2}}=kx+c$$ $$2y^{1/2}=kx+c$$ $$f(x)=\Biggl(\frac{kx+c}{2}\Biggr)^2$$ Plug in the constants and you're good to go.

Edit: For the second one: $$f(a)=b$$ $$\Biggl(\frac{ka+c}{2}\Biggr)^2=b$$ $$b^{1/2}=\frac{ka+c}{2}$$ $$ka+c=2\sqrt{b}$$ $$c=2\sqrt{b}-ka$$

Answer 2

$$\int y^{-\frac12} \, dy = -0.04 \int \, dt $$

$$\frac{\sqrt{y}}{\frac12}=-0.04t+c$$

$$y=\left(\frac{-0.04t+c}{2}\right)^2$$

when $t=0$ , we have $y(0)=\left(\frac{c}2\right)^2$, that is $c = \pm 2\sqrt{y(0)} $