MathematicsHard38×since 2002Q3560Let y=y(x),y>0y=y(x), y > 0y=y(x),y>0, be a solution curve of the differential equation (1+x2)dy=y(x−y)dx\left(1+x^{2}\right) \mathrm{d} y=y(x-y) \mathrm{d} x(1+x2)dy=y(x−y)dx. If y(0)=1y(0)=1y(0)=1 and y(22)=βy(2 \sqrt{2})=\betay(22)=β, thenAeβ−1=e−2(3+22)e^{\beta^{-1}}=e^{-2}(3+2 \sqrt{2})eβ−1=e−2(3+22)Be3β−1=e(5+2)e^{3 \beta^{-1}}=e(5+\sqrt{2})e3β−1=e(5+2)Ce3β−1=e(3+22)e^{3 \beta^{-1}}=e(3+2 \sqrt{2})e3β−1=e(3+22)Deβ−1=e−2(5+2)e^{\beta^{-1}}=e^{-2}(5+\sqrt{2})eβ−1=e−2(5+2)Check answerSkip