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Find the general solution to these differential equations dy/dx=y-3
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@agent0smith
the answer was \[y=Ce^x + 3\]
\[\frac{ dy}{ dx}=y-3 = 1(y-3)\] \[\frac{ dy }{ y-3 } = 1 dx\] Now integrate both sides, then you'll have to take the power of e on both sides.
\[\frac{ 1 }{ y-3 }=x+c\]
did i integrate it right?
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on the left hand side?
You might want to try a substitution on the right... let u = x-3, then the integral of 1/u is ln u
wait nvm i got it lol
can you help me with another one please? <3
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