Question

Decide a solution to a differential equation 3y' -15y=0 that satisfies y'(0)=2
Thank you.

Answer 1

auxiliary equation: 3r-15=0, r=5
general solution: y=a*e^(5x)
y'(x)=5a*e^(5x)
y'(0)=5a=2, a=0.4
therefore solution is y=0.4*e^(5x)

Answer 2

or like this:
\[3\frac{ dy }{ dx }=15y \Rightarrow \frac{dy }{ y }=5dx\]
integrate both sides
\[\int\limits \frac{ dy }{ y}=5\int\limits dx\]
\[lny=5x+C\]
exponentiate:
\[y=Ce^{5x} \Rightarrow y'=5Ce^{5x}\]
I think you can finish this out now

Answer 3

you can solve like this also.
3y'-15y=0
ory'-5y=0
when x=0,y'=2
2-5y=0
y=2/5
again y' =5y
or dy/dx=5y
dy/y=5dx
\[\int\limits \frac{ dy }{y }=5\int\limits dx\]
\[\ln y =5x+c\]
\[y= e ^{5x+c}=e ^{c}e ^{5x}=Ce ^{5x}\]
when x=0 ,y=2/5
\[\frac{ 2 }{5 }=C e ^{0}=C\]
\[y=\frac{ 2 }{5 }e ^{5x}=0.4 e ^{5x}\]

Answer 4

thank you all for the reply