Let x=x(y) be the solution of the differential equation 2(y+2) _e(y+2)dx+(x+4-2 _e(y+2))dy…

Question

Let $x=x(y)$ be the solution of the differential equation

$2(y+2){\log }_{e}(y+2)dx+(x+4-2{\log }_{e}(y+2))dy=0,y>-1$

with $x({e}^{4}-2)=1$. Then $x({e}^{9}-2)$ is equal to :

Accepted Answer

$\begin{gathered} 2(y+2)\ln (y+2)dx+(x+4-2\ln (y+2))dy=0 \ 2\ln (y+2)+(x+4-2\ln (y+2))\frac{1}{y+2}\cdot \frac{dy}{dx}=0 \ \ln (y+2)=t \ \frac{1}{y+2}\cdot \frac{dy}{dx}=\frac{dt}{dx} \ 2t+(x+4-2t)\cdot \frac{dt}{dx}=0 \ (x+4-2t)\frac{dt}{dx}=-2t \ \frac{dx}{dt}=\frac{2t-4-x}{2t} \ \frac{dx}{dt}+\frac{x}{2t}=\frac{2t-4}{2t} \end{gathered}$

$\begin{gathered} x\cdot {t}^{1/2}=\int \frac{2t-4}{2t}\cdot {t}^{1/2}\cdot dt \ x\cdot {t}^{1/2}=\int ({t}^{1/2}-\frac{2}{{t}^{1/2}})\cdot dt \ =\frac{{t}^{\frac{3}{2}}}{\frac{3}{2}}-2\cdot \frac{{t}^{\frac{1}{2}}}{\frac{1}{2}}+C \ x\cdot {t}^{\frac{1}{2}}=\frac{2{t}^{\frac{3}{2}}}{3}-4{t}^{\frac{1}{2}}+C \ x=\frac{2}{3}\cdot t-4+C\cdot {t}^{\frac{-1}{2}} \ x=\frac{2}{3}\ln (y+2)-4+C\cdot (\ln (y+2){)}^{\frac{-1}{2}} \end{gathered}$

Put $y={e}^{4}-2,x=1$

$\begin{gathered} 1=\frac{2}{3}	imes 4-4+C	imes \frac{1}{2} \ \frac{C}{2}=5-\frac{8}{3}=\frac{7}{3} \ C=\frac{14}{3} \ x=\frac{2}{3}	imes 9-4+\frac{14}{3}	imes \frac{1}{3} \ =2+\frac{14}{9} \ =\frac{32}{9} \end{gathered}$

Solutions · 1