Question

please help me answer this differentiation question

Answer 1

y=e^(4*x^2)

diffrentiate y

dy/dx=(4x^2)*8*x*e^(4*x^2)

d2y/dx2=(4x^2)*(8*x)*e^(4*x^2)*8*x*4*x^2*e^(4*x^2)+8*x*8*x*e^(4*x^2)+4*x^2*8*e^(4*x^2)

now substitute in the equation to prove it

Answer 2

Hence proved that $y=e^{4x^{2} }$ satisfies the $\frac{1}{4} \frac{d^{2} y}{dx^{2} }-2x\frac{dy}{dx} -2y=0$ .

Step-by-step explanation:

We are given,

$y=e^{4x^{2} }$  

and we have to prove  

$\frac{1}{4} \frac{d^{2} y}{dx^{2} }-2x\frac{dy}{dx} -2y=0$ .

• Formula used,

Differentiation of exponential term if $y=e^{ax}$ ,

1. $\frac{dy}{dx} =e^{ax}.\frac{d}{dy} (ax)$
2. $\frac{d}{dx} (A*B)=A*\frac{d}{dx} B+B*\frac{d}{dx} A$
3. $\frac{d}{dx} (x^{n})=nx^{n-1}$

• Solving given term,

We have $y=e^{4x^{2} }$  

by using formula (1) we get Differentiation of y is,

$\frac{dy}{dx} =e^{4x^{2} } *\frac{d}{dx} (4x^{2} )$

    $=e^{4x^{2} } *4*2(x)^{2-1}$

    $=e^{4x^{2} } *8x$

$\frac{dy}{dx}=8xe^{4x^{2} }$

now by using formula (2) and formula (3) we get the double differentiation,

$\frac{dy}{dx} =e^{4x^{2} } *8x$

$\frac{d^{2} y}{dx^{2} } =e^{4x^{2} } *\frac{d}{dx} (8x)+8x*\frac{d}{dx} e^{4x^{2} }$

     $=e^{4x^{2} } *8*1(x)^{1-1} +8x*(e^{4x^{2} } *4*2(x)^{2-1})$

     $=8e^{4x^{2} } +8x*e^{4x^{2} }*4*2*(x)^{2-1}$

     $=8e^{4x^{2} } +64x^{2} *e^{4x^{2} }$

to prove the question, we take 4 as common from R.H.S and take it to L.H.S,

$\frac{d^{2} y}{dx^{2} } =8e^{4x^{2} } +64x^{2} *e^{4x^{2} }$

     $=4[2e^{4x^{2} } +16x^{2} e^{4x^{2} }]$

$\frac{1}{4} \frac{d^{2} y}{dx^{2} } =2e^{4x^{2} } +16x^{2} e^{4x^{2} }$

we have $y=e^{4x^{2} }$   and  $\frac{dy}{dx}=8xe^{4x^{2} }$  so we can write,

$\frac{1}{4} \frac{d^{2} y}{dx^{2} } =2e^{4x^{2} } +2*8x^{2} e^{4x^{2} }$

$\frac{1}{4} \frac{d^{2} y}{dx^{2} } =2y +2\frac{dy}{dx}$

taking R.H.S to L.H.S,

$\frac{1}{4} \frac{d^{2} y}{dx^{2} } -2y -2\frac{dy}{dx}=0$

in this way we proved our given question.