习题7.3

\paragraph{题目}

求解下列微分方程.

(1) \(y''=2x-\cos x, y(0)=1, y'(0)=-1\);

(2) \(xy''+(y')^2-y'=0, y(1)=1-\ln 2, y'(1)=\frac{1}{2}\);

(3) \(y''=3\sqrt{y}, y(0)=1, y'(0)=2\);

(4) \((1+x^2)y''-2xy'=0\);

(5) \(y''+\frac{2}{1-y}(y')^2=0\);

(6) \((y''')^2+(y'')^2=1\);

(7) \(xy''-y'\ln y'+y'=0\);

(8) \((1+x^2)y''+(y')^2=-1\);

(9) \((y'')^2-y'=0\).

解

(1) 等式左右两边积分, 可得 \(y' = x^2-\sin x + C_1\). 由 \(y'(0)=-1\) 得 \(C_1 = -1\), 故 \(y'=x^2-\sin x - 1\). 等式左右两边再次积分, 可得 \(y=\frac{1}{3}x^3+\cos x - x + C_2\). 由 \(y(0)=1\) 得 \(C_2 = 0\), 故 \(y = \frac{1}{3}x^3 + \cos x - x\).

(2) 设 \(p = y'\), 则有 \(xp' + p^2 - p = 0\), 即 \(\frac{\text{d}{p}}{p - p^2} = \frac{\text{d}{x}}{x}\). 化简为 \(\left(\frac{1}{p} - \frac{1}{p - 1}\right)\text{d}{p} = \frac{\text{d}{x}}{x}\) 后, 积分得 \(\ln\left|\frac{p-1}{p}\right| = \ln \left|x\right| + C_1\), 此即 \(\frac{p - 1}{p} = \frac{C_2}{x}\). 由于 \(p(1) = y'(1) = \frac{1}{2}\), 因此 \(C_2 = -1\). 故 \(p = y' = 1 - \frac{1}{x + 1}\). 再次积分得 \(y = x - \ln(x + 1) + C\). 由于 \(y(1) = 1 - \ln 2\), 故 \(C = 0\). 所以 \(y = x - \ln(x + 1)\).

(3) 设 \(p = y'\), 则有 \(p\frac{\text{d}{p}}{\text{d}{y}} = 3\sqrt{y}\), 即 \(p\text{d}{p} = 3\sqrt{y}\text{d}{y}\). 积分可得 \(\frac{1}{2}p^2=2y^{\frac{3}{2}}+C_1\). 由 \(y'(0) = p(0) = 2\) 得 \(C_1 = 0\), 故 \(y' = p = 2y^{\frac{3}{4}}\)(负值舍去). 此即 \(\frac{\text{d}{y}}{y^{\frac{3}{4}}} = 2\text{d}{x}\). 积分得 \(4\sqrt[4]{y} = 2x + C_2\). 由于 \(y(0) = 1\), 故 \(C_2 = 4\). 所以 \(y = \left(\frac{1}{2}x + 1\right)^4\).

(4) 注意到

\[ \left(\frac{y}{1 + x^2}\right)' = \frac{(1 + x^2)y'' - 2xy'}{(1 + x^2)^2} = 0 \]

因此 \(y' = C_1(1 + x^2)\), 积分得 \(y = C_1(\frac{1}{3}x^3 + x + C)\), 这等价于 \(y = C_1(\frac{1}{3}x^3 + x) + C_2\).

(5) 设 \(p = y'\), 则有 \(p\frac{\text{d}{p}}{\text{d}{y}} + \frac{2}{1-y}p^2=0\). 这个方程可以化简为 \(\frac{\text{d}{p}}{p} = \frac{2\text{d}{y}}{y-1}\). 积分得 \(\ln \left|p\right| = 2\ln\left|y - 1\right| + C_1\) 即 \(\left|p\right| = e^{C_1}(y-1)^2\). 由于 \(p \equiv 0\) 也为方程的解, 故方程的解可表示为 \(p = C_2(y-1)^2\). 因此 \(\frac{\text{d}{y}}{\text{d}{x}} = C_2(y - 1)^2\), 即 \(\frac{\text{d}{y}}{(y - 1)^2} = C_2\text{d}{x}\). 积分得 \(\frac{1}{1 - y} = C_2 x + C_3\). 故原方程的解为 \(y = 1 - \frac{1}{C_2 x + C_3}\).

(6) 设 \(p = y''\), 则有 \((p')^2 + p^2 = 1\), 即 \(\frac{\text{d}{p}}{\sqrt{1 - p^2}} = \pm \text{d}{x}\), 或 \(p^2 \equiv 1\).

对于前者而言, 积分得 \(x = \arcsin p - C_1\), 即 \(y'' = p = \sin(x + C_1)\)(没有正负号是因为可以取 \(C_1\gets C_1 + \pi\) 得到负号). 积分得 \(y' = -\cos(x + C_1) + C_2\), 再积分得 \(y = -\sin(x + C_1) + C_2 x + C_3\).

对于后者而言, 易知有解 \(y = \pm \frac{1}{2}x^2 + C_1 x + C_2\).

(7) 注意到

\[ \left(\frac{\ln y' - 1}{x}\right)' = \frac{\left(\frac{y''}{y'}\right)x-\ln y' + 1}{x^2} = \frac{xy''-y'\ln y' + y'}{x^2y'} = 0 \]

因此 \(\frac{\ln y' - 1}{x} = C_1\), 即 \(\ln y' - 1 = C_1 x\), 化简得 \(\text{d}{y} = e^{C_1 x + 1}\text{d}{x}\). 积分得 \(y = \frac{1}{C_1}e^{C_1 x + 1} + C_2\).

(8) 设 \(p = y'\), 则有 \(1+p^2+p'(1+x)^2=0\), 即 \(\frac{\text{d}{p}}{1+p^2} = -\frac{\text{d}{x}}{1+x^2}\). 积分得 \(\arctan p = -\arctan x + C\), 即 \(\frac{x + p}{1 - xp} = \tan C = C_1\). 化简得 \(p = \frac{C_1 - x}{1 + C_1 x}\), 积分得 \(y = \ln(1 + C_1 x) - \frac{1}{C_1}\cdot \frac{1}{1 + C_1 x} +C_2(C_1 \neq 0)\), 或 \(y = -\frac{1}{2}x^2 + C_2(C_1 = 0)\).

(9) 设 \(p = y'\), 则有 \((p')^2 = p\), 即 \(p' = \pm \sqrt{p}\), 化简得 \(\frac{\text{d}{p}}{\sqrt{p}} = \pm \text{d}{x}\), 或 \(p \equiv 0\).

对于前者而言, 积分得 \(x = \pm 2\sqrt{p} + C_1\), 即 \(y' = p = \frac{1}{4}(x - C_1)^2\), 再次积分得 \(y = \frac{1}{12}(x - C_1)^3 + C_2\).

对于后者而言, 易知有解 \(y = C_3\).