First Order Checking;Show that y=x³+3x is a solution of dy/dx-3x(x+1)=0?
I don't really understand how to get the answer. This is my working so far
differentiate y=x³+3x
becomes dy/dx=3x²+3
Next I sub into dy/dx-3x(x+1)=0
therefore (3x²+3)-(3x²+3x[expanded])=0
but this doesnt equal zero, the answer is 3-3x=0
Help?
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dy/dx-3x(x+1)=0
y=x³+3x
→dy/dx=3x²+3=3(x³+1)
→ dy/dx−3(x³+1) =0
Hence y=x³+3 is not a solution of dy/dx-3x(x+1)=0.
y=x³+3x
=>dy/dx = 3x^2+3
=> dy/dx - 3(x+1) = 0
You are perfectly correct.
The general solution is y=x³ + (3/2)x² + C and
it seems that the setter has made a mistake.