For x ∈ ℝ, let y(x) be a solution of the differential equation $(x^{2}−5)$dy/dx − 2xy $= −2x(x^{2}−5)^{2}$ such that $y(2)=7.$ Then the maximum value of the function y(x) is

For x ∈ R, let y(x) be a solution of the differential equation (x²-5)dy/dx-2xy=-2x(x²-5)² such that y(2)=7. Then the maximum value of the function y(x) is

$(x^2-5)\dfrac{dy}{dx}-2xy=-2x(x^2-5)^2$

$y(2)=7$

$\text{To find: maximum value of } y(x)$

$\dfrac{d}{dx}\left[\dfrac{y}{x^2-5}\right]=\dfrac{(x^2-5)y'-2xy}{(x^2-5)^2}$

$\dfrac{d}{dx}\left[\dfrac{y}{x^2-5}\right]=-2x$

$\dfrac{y}{x^2-5}=-x^2+C$

$y=(x^2-5)(C-x^2)$

$y(2)=(-1)(C-4)=7 \;\Rightarrow\; C-4=-7 \;\Rightarrow\; C=-3$

$y(x)=(x^2-5)(-x^2-3)=-x^4+2x^2+15$

$\text{Plan: } y'=0 \Rightarrow \text{critical points, then } y'' \text{ test}$

$y'=-4x^3+4x=4x(1-x^2)=4x(1-x)(1+x)$

$y'=0 \Rightarrow x=-1,\ 0,\ 1$

$y''=-12x^2+4:\quad y''(0)=4>0,\quad y''(\pm1)=-8<0$

$y(\pm1)=-1+2+15=16$

$x\to\pm\infty:\ y=-x^4+\dots\to-\infty$

$\max y(x)=16$