0

If $x^4+7x^2y^2+9y^4=24xy^3$. Then value of $y''(1,\alpha)$ is ,given $1+7\alpha^2+9\alpha^4=24\alpha^3$

What I try:

First differentiate both side with respect to $x$, we have

$\displaystyle 4x^3+7(x^2+2yy'+y^2\cdot 2x)+36y^3y'=24(x\cdot 3y^2y'+y^3\cdot 1)\cdots \cdots (1) $

Put $x=1,y=\alpha$, we get

$\displaystyle 4+7[1+2\alpha\cdot y'(1,\alpha)+\alpha^2]+36\alpha^3\cdot y'(1,\alpha)=24[3\alpha^2 \cdot y'(1,\alpha)+\alpha^3]$

$\displaystyle y'(1,\alpha)=\frac{24\alpha^3-7\alpha^2-11}{14\alpha+36\alpha^3-72\alpha^2}$

Again differentiate w.r to $x,$ We get

$\displaystyle 12x^2+14x+14[y\cdot y''+(y')^2]+14yy'+36[y^3y''+y'\cdot 3y^2y']=72y^2y'+72[x(y^2y''+2y(y')^2)+y^2y']\cdots \cdots (2)$

Now putting $x=1$ and $y=\alpha$ in $(2)$

And we get value of $y''(1,\alpha)$.

Above method is very lengthy

Can anyone please tell me How to solve in some short way, because I have seems that each term has sum of power of $x$ and $y$ is $4$

Please have a look on that problem, Thanks

jacky
  • 5,194
  • https://math.stackexchange.com/questions/1406677/if-x47x2y29y4-24xy3-show-that-fracdydx-fracyx?rq=1 – jacky Sep 25 '23 at 08:58

4 Answers4

2

Note: the zero set of a homogeneous $F(x,y)$ is a union of lines through origin or just the origin itself.

So $y''=0$ when defined.

user10354138
  • 33,239
1

Hint.

Making $y = a x$ and substituting

$$ \left(9 a^4-24 a^3+7 a^2+1\right) x^4=0 $$

and

$$ p(a)=9 a^4-24 a^3+7 a^2+1 = 9(a-r_1)(a-r_2)((a+c_1)^2+c_2^2)=0 $$

Note that $p(a)=0$ has at least a real root due to the coefficient signs, which leads to the conclusion that there are at least two real roots because the remaining third order polynomial should have at least, a real root. In fact, the polynomial can be solved, obtaining

$$ 9 (a-2.32298) (a-0.543495) \left((a+0.0999063)^2+0.0780254\right)=0 $$

or

$$ 9 \left(\frac yx-2.32298\right) \left(\frac yx-0.543495\right) \left(\left(\frac yx+0.0999063\right)^2+0.0780254\right)=0 $$

so we can follow assuming $(r_1, r_2)\in \mathbb{R}$ and $(a+c_1)^2+c_2^2\gt 0$

As expected due to linear behavior

$$ y''=0 $$

Cesareo
  • 33,252
1

Let $$F(x,y)=x^4+7x^2y^2+9y^4-24xy^3=0.\tag1$$ Using the formula of the derivative of the implicit function, easily to get $$P(x,y)=\dfrac{\text dy}{\text dx} =-\dfrac{\dfrac{\partial F}{\partial x}}{\dfrac{\partial F}{\partial y}} = -\dfrac{4x^3+14xy^2-24y^3}{14x^2y+36y^3-72xy^2}=-\dfrac{2x^3+7xy^2-12y^3}{y(7x^2+18y^2-36xy)}.$$ Then, by the formula of the derivative of the function of some unknowns, $$\dfrac{\text d^2y}{\text dx^2}=\dfrac{\text dP}{\text dx} ={\dfrac{\partial P}{\partial x}}+P\dfrac{\partial P}{\partial y} =\dfrac{-14x^4+144x^3y-59x^2y^2-168xy^3 + 306 y^4}{y(7x^2-36xy+18y^2)^2}+{\dfrac{x(-14x^4+144x^3y-59x^2y^2-168xy^3+306 y^4)}{y^2(7x^2-36xy+18y^2)^2}} \dfrac{2x^3+7xy^2-12y^3}{y(7x^2+18y^2-36xy)}$$ $$=\dfrac{2(-14x^4+144x^3y-59x^2y^2-168xy^3+306y^4)(x^4+7x^2y^2-24xy^3+9y^4)}{y^3(7x^2-36xy+18y^2)^3}.$$ Finally, $$\color{green}{\mathbf{\dfrac{\text d^2y}{\text dx^2}\bigg|_{(1,\alpha)} =\dfrac{2(-14+144\alpha-59\alpha^2-168\alpha^3+306\alpha^4)(1+7\alpha^2-24\alpha^3+9\alpha^4)}{\alpha^3(7-36\alpha+18\alpha^2)^3}.}}$$

Yuri Negometyanov
  • 28,026
1

Let $z=z(x)=\frac{y(x)}x.$ Then $9z^4-24z^3+7z^2+1=0$ and $(36z^3-72z^2+14z)z'=0.$ At the $P=(x,y)=(1,\alpha)$ we have $z=\alpha.$ So if $\color{red}{2\alpha(18\alpha^2-36\alpha+7)\alpha\neq 0}$, we get $z'\vert_P=0$ and $z''\vert_P=0.$

Now, $z'\vert_P=\frac{y'x-y}{x^2}\vert_P$ gives $y'\vert_P=\alpha.$

Next, $z''\vert_P=\frac{x^2y''-2xy'+2y}{x^3}\vert_P$ gives $y''\vert_P=0.$

Bob Dobbs
  • 10,988