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

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).

• $c$ may not necessarily be equal to $y(0)$. If you plug in $t=0$, you get an equation for $c$ in terms of $y(0)$, though. – Matthew Leingang Oct 16 '18 at 1:59

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

• I needed the second equation you did. So I would solve for c, and since im given that y(0)=3, I would end up with 2sqrt(3) = c? – Jytrex Oct 16 '18 at 3:36
• @Jytrex How about now? – clathratus Oct 16 '18 at 4:14
• yes thanks! so if I were to do f(0) =3, ka would = 0, and then have 2sqrt(3) – Jytrex Oct 16 '18 at 18:11
• @Jytrex If the solution helps, give it a big 'ole check – clathratus Oct 16 '18 at 22:59
• sure, and so I am correct in saying the value for c? – Jytrex Oct 17 '18 at 2:15

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

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

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

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

• How do I know whether to use + or minus? I was given y(0) = 3, and the equation represents water leaving a tank. – Jytrex Oct 16 '18 at 2:06
• @Jytrex Both are possible according to Wolfram Alpha. – Toby Mak Oct 16 '18 at 2:11
• From the second formula you get uniquely $c=2\sqrt{y_0}+0.04t_0$. Note that for $t\ge25c$ you get $y\equiv0$. Any solution $y$ has to be falling by the sign of $y'$ in the ODE. – LutzL Oct 16 '18 at 5:54