If frac{d^n y}{d x^n}=y_n and y=e^{sqrt{x}}+e | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $y = e^{\sqrt{x}} + e^{-\sqrt{x}}$.First derivative $y_1 = \frac{d}{dx}(e^{\sqrt{x}} + e^{-\sqrt{x}}) = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x

Vedclass · Accepted Answer

$y$

Vedclass · Answer

(B) Given $y = e^{\sqrt{x}} + e^{-\sqrt{x}}$.First derivative $y_1 = \frac{d}{dx}(e^{\sqrt{x}} + e^{-\sqrt{x}}) = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} - e^{-\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} = \frac{e^{\sqrt{x}} - e^{-\sqrt{x}}}{2\sqrt{x}}$.Multiplying by $2\sqrt{x}$,we get $2\sqrt{x} y_1 = e^{\sqrt{x}} - e^{-\sqrt{x}}$.Differentiating both sides with respect to $x$:$2\sqrt{x} y_2 + 2 y_1 \cdot \frac{1}{2\sqrt{x}} = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} + e^{-\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}$.$2\sqrt{x} y_2 + \frac{y_1}{\sqrt{x}} = \frac{e^{\sqrt{x}} + e^{-\sqrt{x}}}{2\sqrt{x}} = \frac{y}{2\sqrt{x}}$.Multiplying the entire equation by $2\sqrt{x}$:$4 x y_2 + 2 y_1 = y$.

If $\frac{d^n y}{d x^n}=y_n$ and $y=e^{\sqrt{x}}+e^{-\sqrt{x}}$,then $4 x y_2+2 y_1=$

Similar Questions

If $y = a \cos(\log x) + b \sin(\log x)$ where $a, b$ are parameters,then ${x^2}y'' + xy' = $

If $y=x \log \left(\frac{1}{a x}+\frac{1}{a}\right)$,then $x(x+1) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}-y=$

Let $y = \log_8 \left( \frac{1-x^2}{1+x^2} \right)$ for $-1 < x < 1$. Then at $x = \frac{1}{2}$,the value of $225(y' - y'')$ is equal to:

If $y = \tan(\log x)$,then $\frac{d^2 y}{d x^2} =$

$f(x)$ is a twice differentiable function such that $f^{\prime \prime}(x)=-f(x)$ and $f^{\prime}(x)=g(x)$. If $h(x)=(f(x))^2+(g(x))^2$ and $h(1)=2$, then $h(2)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module