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We have the equation $$\ln(1+2y)+\sin(x^2y)=2$$

find $$y''(0)$$

I think I'm supposed to use implicit differentiation two times, but when I do that I get a really long and confusing answer.

The first time I used implicit differentiation I got $$\frac{dy}{dx}=-\frac{2xy (1+2y) \cos(x^2y)}{2+x^2 (1+2y)\cos(x^2y)}$$

I would appreciate it if someone could help me :)

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  • $\begingroup$ Plug $y=a+bx+cx^2+o(x^2)$ into your equation and identify the coefficients. $\endgroup$ Oct 12, 2022 at 10:47
  • $\begingroup$ There is no need to rearrange for $dy/dx$ as that would necessitate the use of quotient rule. Instead you can implicitly differentiate twice in a row. However since you have this form already you can use it to figure out what $y'(0)$ is. $\endgroup$ Oct 12, 2022 at 10:48

2 Answers 2

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Okay so we have the equation

$$\ln(1+2y(x))+\sin(x^2y(x))=2$$

and we wish to find $y''(0)$. So differentiating once we get that

$$\frac{2y'(x)}{1+2y(x)}+(2xy(x)+x^2y'(x))\cos(x^2y(x))=0.$$

Differentiating again we get

$$\frac{2y''(x)(1+2y(x))-4y'(x)^2}{(1+2y(x))^2}+(2y(x)+4xy'(x)+x^2y''(x))\cos(x^2y(x))-(2xy(x)+x^2y'(x))^2\sin(x^2y(x))=0.$$

Plugging $x=0$ into this we get the equation

$$\frac{2y''(0)(1+2y(0))-4y'(0)^2}{(1+2y(0))^2}+2y(0)=0.$$

This can be easily rearranged to get that

$$y''(0)=\frac{2y'(0)^2}{1+2y(0)}-y(0)(1+2y(0)).$$

To compute $y''(0)$, we thus need to first also find $y'(0)$ and $y(0)$. From our first equation, plugging in $x=0$, we have that

$$\ln(1+2y(0))=2,$$

and so

$$y(0)=\frac{e^2-1}{2}.$$

Similarly, plugging in $x=0$ into our second equation we have that

$$2y'(0)=0,$$

i.e.

$$y'(0)=0.$$

Combining these results we now finally have that

$$y''(0)=-\frac{e^2(e^2-1)}{2}.$$

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You're on the right track... If you substitute $x=0$ in the original equations, you get $y= \frac{e^2-1}{2}$. This means that $y(0)=\frac{e^2-1}{2}$. Using the expression you obtained for $y'$, you can also get $y'(0) = 0$. Now, you just need to differentiate again , substitute $x = 0$, $y = \frac{e^2-1}{2}$, $y'=0$ and solve with respect to $y''$. The first differentiation leads to $$ \frac{2 y'}{1+2y} + (2xy+x^2y')\cos(x^2y)=0 $$ and the second differentiation leads to $$ \dfrac{2y''(1+2y)-2y'\cdot 2y'}{(1+2y)^2} + (2y+4xy'+x^2y'')\cos(x^2y)-(2xy+x^2y')^2 \sin(x^2 y) = 0 $$

If you substitute the known values befiore you solve with respect to $y''$, you get: $$ \frac{2y''(0)}{1+2y(0)} + 2y(0) = 0 \Rightarrow y''(0) = -y(0)(1+2y(0)) = \frac{1}{2} e^2 \left(1-e^2\right) $$

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