Solve dy/dx=y(2x+3)?
Answers:
1/y dy = 2x+3 dx
∫1/y dy = ∫2x+3 dx
ln y = x²+3x+C
y=Ce^(x²+3x)
Unfortunatelly, your initial value is underdefined, since you don't tell us as what x-value y=1. If you mean y(0)=1, then:
1=Ce^(0)
C=1
y=e^(x²+3x)
The equation can be written as
dy/y=(2x+3)dx
integrating both sides we get
ln(y)=x^2+3x or y=e^(x^2+3x)
now putting y=1 we get x=0,-3
Hence, proved
dy/dx=y(2x+3)
dy/y=(2x+3)dx
integrate both sides
lny= x^2+3x+C
>>>> y= e^C.e^(x^2+3x) =Ke^(x^2+3x)
where K and C are constants
we are only told y=1 and not a
corresponding value for x,so the value
of K cannot be found
i hope that this helps
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