题目
(19) (本题满分12分)已知y(x)满足x^2y^n+xy'-9y=0,满足y(1)=2,y'(1)=6(1)利用变换x=e^t将上述方程化为常系数线性方程，并求y(x);(2)计算int_(1)^2y(x)sqrt(4-x^2)dx
(19) (本题满分12分)已知y(x)满足$x^{2}y^{n}+xy'-9y=0$,满足y(1)=2,y'(1)=6
(1)利用变换$x=e^{t}$将上述方程化为常系数线性方程，并求y(x);
(2)计算$\int_{1}^{2}y(x)\sqrt{4-x^{2}}dx$
题目解答
答案
(1) **变换与求解**
令 $x = e^t$，则 $y' = \frac{1}{x} \frac{dy}{dt}$，$y'' = \frac{1}{x^2} \left( \frac{d^2y}{dt^2} - \frac{dy}{dt} \right)$。代入原方程得：
\[
\frac{d^2y}{dt^2} - 9y = 0
\]
特征根为 $r = \pm 3$，通解为：
\[
y(t) = C_1 e^{-3t} + C_2 e^{3t}
\]
转换回 $x$：
\[
y(x) = \frac{C_1}{x^3} + C_2 x^3
\]
由初始条件 $y(1) = 2$，$y'(1) = 6$，解得 $C_1 = 0$，$C_2 = 2$，故：
\[
\boxed{y(x) = 2x^3}
\]
(2) **计算积分**
令 $x = 2\sin\theta$，则：
\[
\int_{1}^{2} 2x^3 \sqrt{4-x^2} \, dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} 64 \sin^3\theta \cos^2\theta \, d\theta = \frac{22\sqrt{3}}{5}
\]
**答案**
\[
\boxed{\frac{22\sqrt{3}}{5}}
\]