Find a solution of the differential equation dy/dx=3y+1 that satisfies the initial condition y(0)=2
I got the answer : y=(7e^(-x)-1)/(3) . Am I right?

Question

Find a solution of the differential equation dy/dx=3y+1 that satisfies the initial condition y(0)=2 
I got the answer : y=(7e^(-x)-1)/(3) . Am I right?

turingtest · Answer

yes

turingtest · Answer

oh no, you forgot the 3 in the exponent

anonymous · Answer

this is what I get$$y = \frac{7e^{3x} - 1}{3}$$

turingtest · Answer

right :)

anonymous · Answer

ops..
can you show me how does it go?

turingtest · Answer

what did you use as an integrating factor?

anonymous · Answer

separate the variables$$\frac{dy}{3y + 1} = dx$$ integrate both sides$$\frac{1}{3}\ln (3y + 1) = x + C$$solve for y$$y = \frac{e^{3x + 3C} - 1}{3}$$rewrite the constant$$y = \frac{De^{3x}-1}{3}$$apply the initial condition and solve for D

turingtest · Answer

oh I didn't even notice it was separable
I'm tired, g'night :)

anonymous · Answer

good night and thank both of you guys~
Im finding my mistake now 0.0