I'm new to Differential equations: ty' = 4y , y(1) = 2 t > 0

[scrum]

I don't quite get how these work.

I have ty' = 4y , y(1) = 2 t > 0

so I used separation of variables

t (dy)/(dt) = 4y

t(dy) = 4y(dt)

(1/4y) (dy) = (1/t) (dt)

integrate both sides

log(4y) = log(t) + C

4y = ct

y = (ct)/4

so from the original thing y(1) = 2

2 = c1/4

so c = 8 and the answer is

y(t) = 2t ?

(it's not, the homework submission computer just told me incorrect)

 

[galactus]

Here's how I done it:

\(\displaystyle ty'=4y, \;\ y(1)=2\)

Separate variables:

\(\displaystyle \frac{y'}{4y}=\frac{1}{t}\)

Integrate:

\(\displaystyle \int\frac{1}{4y}dy=\int\frac{1}{t}dt\)

\(\displaystyle \frac{ln(y)}{4}=ln(t)+C\)

e to both sides:

\(\displaystyle y^{\frac{1}{4}}=e^{c}t\)

\(\displaystyle y=(te^{c})^{4}\)

Now, using the IC:

\(\displaystyle 2=((1)e^{c})^{4}\)

\(\displaystyle c=\frac{ln(2)}{4}\)

\(\displaystyle \boxed{y=2t^{4}}\)

Try that and see if the computer takes it. Plugging this into the original we see it works.

 

[soroban]

Hello, scrum!

A slightly different approach . . .

\(\displaystyle t\!\cdot\!\frac{dy}{dx} \:=\:4y,\quad y(1) \,=\, 2,\;\;t > 0\)

\(\displaystyle \text{Separate variables: }\;\frac{dy}{y} \:=\:4\,\frac{dt}{t}\)

\(\displaystyle \text{Integrate: }\;\ln(y) \;=\;4\ln(t) + c\quad\Rightarrow\quad \ln(y) \;=\;\ln(t^4) + \ln(C)\)

. . \(\displaystyle \ln(y) \;=\;\ln(Ct^4) \quad\Rightarrow\quad y \;=\;Ct^4\)

\(\displaystyle \text{Since }y(1)\,=\,2\text{, we have: }\;2 \;=\;C\!\cdot\!1^4 \quad\Rightarrow\quad C \,=\,2\)


\(\displaystyle \text{Therefore: }\;\boxed{y \;=\;2t^4}\)