Differentiation problem

Question

Prove that 

Y=H^(7x2 +3x +2)

Is the solution of the differentiation equation

(14x+3) Y' - Y'' = 14Y

Answer 1

I think you missed the minus sign on the right hand side. It should be (14x+3) y' - y'' = -14y

 

Also, the H should be Euler's number "e".

 

Solution: Just plug it in.

 

y = e^{7x^2+3x +2} ⇒ y' = (14x+3) e^{7x^2+3x +2}

 

⇒ y'' = [(14x+3)²+14]e^{7x^2+3x +2}=(196x²+84x+23)e^{7x^2+3x+2}

 

(14x+3)y' - y'' = (14x+3)²e^{7x^2+3x+2} - [(14x+3)2+14]e^{7x^2+3x+2} = -14e^{7x^2+3x+2} = -14y

 

 

Answer 2

Do you mean e instead of H?  Should 14y be -14y?

 

y = e^{7x^2+3x+2}

 

y' = e^{7x^2+3x+2}(14x+3)

 

y" = e^{7x^2+3x+2}(14x+3)² + 14e^{7x^2+3x+2}

 

(14x+3)y' - y"=(14x+3)²e^{7x^2+3x+2} - [(14x+3)²e^{7x^2+3x+2}+14e^{7x^2+3x+2}]

 

                   = -14e^{7x^2+3x+2}

 

                   = -14y

 

So, if y = e^{7x^2+3x+2}, then (14x+3)y' - y" = -14y.