laminarskq2p

2023-02-21

How to find ((d^2)y)/(dx^2)?

Camuccinirk84

Beginner2023-02-22Added 4 answers

Defining

$f(x,y\left(x\right))=2xy\left(x\right)+2y{\left(x\right)}^{2}-13=0$ then

$\frac{df}{dx}=2y+2xy\prime +4yy\prime =0$ and

$\frac{d}{dx}\frac{df}{dx}=4y\prime +2xy\prime \prime +4y{\prime}^{2}+4yy\prime \prime =0$

after substitution of $y\prime$ we have

$y\prime \prime =\frac{2xy+2{y}^{2}}{{(x+2y)}^{3}}=\frac{13}{{(x+2y)}^{3}}$

but $y=\frac{1}{2}(-x\pm \sqrt{26+{x}^{2}})$ so finally

$y\prime \prime =\pm \frac{13}{{(26+{x}^{2})}^{\frac{3}{2}}}$

$f(x,y\left(x\right))=2xy\left(x\right)+2y{\left(x\right)}^{2}-13=0$ then

$\frac{df}{dx}=2y+2xy\prime +4yy\prime =0$ and

$\frac{d}{dx}\frac{df}{dx}=4y\prime +2xy\prime \prime +4y{\prime}^{2}+4yy\prime \prime =0$

after substitution of $y\prime$ we have

$y\prime \prime =\frac{2xy+2{y}^{2}}{{(x+2y)}^{3}}=\frac{13}{{(x+2y)}^{3}}$

but $y=\frac{1}{2}(-x\pm \sqrt{26+{x}^{2}})$ so finally

$y\prime \prime =\pm \frac{13}{{(26+{x}^{2})}^{\frac{3}{2}}}$

Ruby Rollins

Beginner2023-02-23Added 5 answers

$2xy+2{y}^{2}=13$

Differentiating wrt $x$ and applying the product rule gives us:

$2\{\left(x\right)\left(\frac{dy}{dx}\right)+\left(1\right)\left(y\right)\}+4y\frac{dy}{dx}=0$

$x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{y}{x+2y}$

Differentiating again wrt $x$ and applying the product rule (twice) gives us:

$\therefore \{\left(x\right)\left(\frac{{d}^{2}y}{{dx}^{2}}\right)+\left(1\right)\left(\frac{dy}{dx}\right)\}+\frac{dy}{dx}+2\{\left(y\right)\left(\frac{{d}^{2}y}{{dx}^{2}}\right)+\left(2\frac{dy}{dx}\right)\left(\frac{dy}{dx}\right)\}=0$

$\therefore x\frac{{d}^{2}y}{{dx}^{2}}+\frac{dy}{dx}+\frac{dy}{dx}+2y\frac{{d}^{2}y}{{dx}^{2}}+2{\left(\frac{dy}{dx}\right)}^{2}=0$

$\therefore x\frac{{d}^{2}y}{{dx}^{2}}+2\frac{dy}{dx}+2y\frac{{d}^{2}y}{{dx}^{2}}+2{\left(\frac{dy}{dx}\right)}^{2}=0$

$\therefore (x+2y)\frac{{d}^{2}y}{{dx}^{2}}+2(-\frac{y}{x+2y})+2{(-\frac{y}{x+2y})}^{2}=0$

$\therefore (x+2y)\frac{{d}^{2}y}{{dx}^{2}}=\frac{2y}{x+2y}-\frac{2{y}^{2}}{{(x+2y)}^{2}}$

$\therefore \frac{{d}^{2}y}{{dx}^{2}}=\frac{2y}{{(x+2y)}^{2}}-\frac{2{y}^{2}}{{(x+2y)}^{3}}$

Overlaying a common denominator results in:

$\frac{{d}^{2}y}{{dx}^{2}}=\frac{\left(2y\right)(x+2y)-\left(2{y}^{2}\right)}{{(x+2y)}^{3}}$

$\therefore \frac{{d}^{2}y}{{dx}^{2}}=\frac{(2xy+4{y}^{2}-2{y}^{2})}{{(x+2y)}^{3}}$

$\therefore \frac{{d}^{2}y}{{dx}^{2}}=\frac{(2xy+2{y}^{2})}{{(x+2y)}^{3}}$

$\therefore \frac{{d}^{2}y}{{dx}^{2}}=\frac{13}{{(x+2y)}^{3}}$ (as $2xy+2{y}^{2}=13$)

Differentiating wrt $x$ and applying the product rule gives us:

$2\{\left(x\right)\left(\frac{dy}{dx}\right)+\left(1\right)\left(y\right)\}+4y\frac{dy}{dx}=0$

$x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{y}{x+2y}$

Differentiating again wrt $x$ and applying the product rule (twice) gives us:

$\therefore \{\left(x\right)\left(\frac{{d}^{2}y}{{dx}^{2}}\right)+\left(1\right)\left(\frac{dy}{dx}\right)\}+\frac{dy}{dx}+2\{\left(y\right)\left(\frac{{d}^{2}y}{{dx}^{2}}\right)+\left(2\frac{dy}{dx}\right)\left(\frac{dy}{dx}\right)\}=0$

$\therefore x\frac{{d}^{2}y}{{dx}^{2}}+\frac{dy}{dx}+\frac{dy}{dx}+2y\frac{{d}^{2}y}{{dx}^{2}}+2{\left(\frac{dy}{dx}\right)}^{2}=0$

$\therefore x\frac{{d}^{2}y}{{dx}^{2}}+2\frac{dy}{dx}+2y\frac{{d}^{2}y}{{dx}^{2}}+2{\left(\frac{dy}{dx}\right)}^{2}=0$

$\therefore (x+2y)\frac{{d}^{2}y}{{dx}^{2}}+2(-\frac{y}{x+2y})+2{(-\frac{y}{x+2y})}^{2}=0$

$\therefore (x+2y)\frac{{d}^{2}y}{{dx}^{2}}=\frac{2y}{x+2y}-\frac{2{y}^{2}}{{(x+2y)}^{2}}$

$\therefore \frac{{d}^{2}y}{{dx}^{2}}=\frac{2y}{{(x+2y)}^{2}}-\frac{2{y}^{2}}{{(x+2y)}^{3}}$

Overlaying a common denominator results in:

$\frac{{d}^{2}y}{{dx}^{2}}=\frac{\left(2y\right)(x+2y)-\left(2{y}^{2}\right)}{{(x+2y)}^{3}}$

$\therefore \frac{{d}^{2}y}{{dx}^{2}}=\frac{(2xy+4{y}^{2}-2{y}^{2})}{{(x+2y)}^{3}}$

$\therefore \frac{{d}^{2}y}{{dx}^{2}}=\frac{(2xy+2{y}^{2})}{{(x+2y)}^{3}}$

$\therefore \frac{{d}^{2}y}{{dx}^{2}}=\frac{13}{{(x+2y)}^{3}}$ (as $2xy+2{y}^{2}=13$)

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