If the solution curve y=f(x) of the differential equation (x^2-4)y^'-2xy+2x(4-x^2)^2=0,x2…

Question

If the solution curve $y=f(x)$ of the differential equation

$({x}^{2}-4){y}^{'}-2xy+2x{(4-{x}^{2})}^{2}=0,x>2,$

passes through the point $(3,15)$, then the local maximum value of $f$ is $____$

Answer: 16

Accepted Answer

$({x}^{2}-4){y}^{'}-2xy+2x{(4-{x}^{2})}^{2}=0,x>2$

rearranging terms.

$\begin{gathered} ({x}^{2}-4)\frac{dy}{dx}-2xy=-2x{({x}^{2}-4)}^{2} \ \frac{dy}{dx}-\frac{2x}{{x}^{2}-4}y=-2x({x}^{2}-4) \end{gathered}$

this is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$

$P=-\frac{2x}{{x}^{2}-4},Q=-2x({x}^{2}-4)$

finding integrating factor (I.F.).

$\begin{gathered} ={e}^{\int Pdx}={e}^{\int -\frac{2x}{{x}^{2}-4}dx} \ ={e}^{-\ln ({x}^{2}-4)}=\frac{1}{{x}^{2}-4} \end{gathered}$

the solution is $y\cdot (I.F)=.\int Q\cdot (I.F)dx+$.

$\begin{gathered} y\cdot \frac{1}{{x}^{2}-4}=\int -2x({x}^{2}-4)\cdot \frac{1}{{x}^{2}-4}dx+C \ \frac{y}{{x}^{2}-4}=\int -2xdx+C \ \frac{y}{{x}^{2}-4}=-{x}^{2}+C\Rightarrow y=({x}^{2}-4)(C-{x}^{2}) \end{gathered}$

substitute point $(3,15)$

$15=({3}^{2}-4)(C-{3}^{2})$

$\begin{gathered} 15=5(C-9)\Rightarrow 3=C-9\Rightarrow C=12 \ f(x)=({x}^{2}-4)(12-{x}^{2}) \ f(x)=-{x}^{4}+16{x}^{2}-48 \end{gathered}$

finding critical points for local maximum.

$\begin{gathered} {f}^{'}(x)=-4{x}^{3}+32x \ {f}^{'}(x)=0\Rightarrow -4x({x}^{2}-8)=0 \ x>2,{x}^{2}=8\Rightarrow x=\sqrt{8}=2\sqrt{2} \end{gathered}$

${f}^{'}((2\sqrt{2}{)}^{-})=$positive, ${f}^{'}((2\sqrt{2}{)}^{+})=$negative

so sign of ${f}^{'}(x)$ changing positive to negative about $x=2\sqrt{2}$ so $x=2\sqrt{2}$ is point of local maxima

calculating local maximum value.

$\begin{gathered} f(2\sqrt{2})=((2\sqrt{2}{)}^{2}-4)(12-(2\sqrt{2}{)}^{2}) \ =(8-4)(12-8) \ =4	imes 4=16 \end{gathered}$

the local maximum value is 16 .

Solutions · 1