一个坑爹题目

战巡

据说是某文科数学的题目...........

已知函数$y=f(x)$在$x\ge 1$上连续且可积，$x=1, x=t, y=0, y=f(x)$围成一个曲边梯形，且对于任意$t>1$，都有这个曲边梯形围绕$x$轴旋转一周得到的旋转体体积为曲边梯形面积的$\pi t$倍
求$f(x)$

isee

怎么理解这个t>1?

战巡

回复 2# isee


就是一个变动的上限啊

爪机专用

\[\int_1^t\pi f(x)^2 dx=t\pi \int_1^t f(x)dx\]
\[f(t)^2=\int_1^t f(x)dx+tf(t)\]
\[2f(t)f'(t)=f(t)+f(t)+tf'(t)\]
不会解。。。

icesheep

把微分方程看成是 x=x(y) 就能解了

战巡

回复 5# icesheep

解这个玩意对我们来说不会很难，问题是这是文科数学，尼玛那帮人没学过微分方程..........
反正我就不知道为啥会出现这样的题目

kuing

在哪看到的啊……

战巡

回复 7# kuing


被人问到的呃..............

isee

哈哈，原来如此~

kuing

求出处……看看是哪位神银出嘀……

hbghlyj

爪机专用 发表于 2014-4-11 04:07
\begin{array}l
2f(t)f'(t)=f(t)+f(t)+tf'(t)\\
f(1)=0\end{array}

用DSolve

1. DSolve[{2 f[t] f'[t] - 2 f[t] == t f'[t], f[1] == 0}, f[t], t]

Copy the Code

解得$f$恒等于0啊

---

用WolframAlpha Get LaTeX form from step by step solution
first-order nonlinear ordinary differential equation
两种方法:
Solve as a homogeneous equation:

1. content=WolframAlpha["2f(t)f'(t)=f(t)+f(t)+tf'(t),f(1)=0",{{"Solution",2},"Content"},PodStates->{"Solution__Step-by-step solution"}]

Copy the Code

\begin{array}{l}
\text{Solve }2 f(t) \frac{df(t)}{dt}=t \frac{df(t)}{dt}+2 f(t)\text{, such that }f(1)=0: \\
\text{Let }f(t)=t v(t), \text{which gives }\frac{df(t)}{dt}=t \frac{dv(t)}{dt}+v(t): \\
2 t \left(t \frac{dv(t)}{dt}+v(t)\right) v(t)=t \left(t \frac{dv(t)}{dt}+v(t)\right)+2 t v(t) \\
\text{Simplify:} \\
2 t \left(t \frac{dv(t)}{dt}+v(t)\right) v(t)=t \left(t \frac{dv(t)}{dt}+3 v(t)\right) \\
\text{Solve for }\frac{dv(t)}{dt}: \\
\frac{dv(t)}{dt}=\frac{-2 v(t)^2+3 v(t)}{t (2 v(t)-1)} \\
\text{Divide both sides by }\frac{-2 v(t)^2+3 v(t)}{2 v(t)-1}: \\
\frac{\frac{dv(t)}{dt} (2 v(t)-1)}{-2 v(t)^2+3 v(t)} =\frac{1}{t} \\
\text{Integrate both sides with respect to }t: \\
\int \frac{\frac{dv(t)}{dt} (2 v(t)-1)}{-2 v(t)^2+3 v(t)} \, dt =\int \frac{1}{t} \, dt \\
\text{Evaluate the integrals:} \\
-\frac{2}{3} \log (-2 v(t)+3)-\frac{1}{3} \log (v(t)) =\log (t)+c_1\text{, where }c_1\text{ is an arbitrary constant.} \\
\text{Substitute back for }f(t)=t v(t): \\
\text{Answer:}\\
-\frac{1}{3} \log \left(\frac{f(t)}{t}\right)-\frac{2}{3} \log \left(-\frac{2 f(t)}{t}+3\right)=\log (t)+c_1 \\
\end{array}
Transform into an exact equation:

1. content=WolframAlpha["2f(t)f'(t)=f(t)+f(t)+tf'(t),f(1)=0",{{"Solution",2},"Content"},PodStates->{"Solution__Step-by-step solution","Solution__Transform into an exact equation"}]

Copy the Code

\begin{array}{l}
\text{Solve }2 f(t) \frac{df(t)}{dt}=t \frac{df(t)}{dt}+2 f(t)\text{, such that }f(1)=0: \\
\text{Rewrite the equation:} \\
-2 f(t)+(-t+2 f(t)) \frac{df(t)}{dt}=0 \\
\text{Let }R(t,f)=-2 f \text{ and }S(t,f)=2 f-t. \\
\text{This is not an exact equation, because }\left(\frac{\partial R(t,f)}{\partial f}=-2\right)\neq \left(-1=\frac{\partial S(t,f)}{\partial
  t}\right). \\
\text{Find an integrating factor }\mu (f) \text{ such that }\mu (f) R(t,f)+\frac{df(t)}{dt} \mu (f) S(t,f)=0 \text{ is exact.} \\
\text{This means }\frac{\partial \text{}}{\partial f}(\mu (f) R(t,f))=\frac{\partial \text{}}{\partial t}(\mu (f) S(t,f)): \\
-2 f \frac{d\mu (f)}{df}-2 \mu (f) =-\mu (f) \\
\text{Isolate }\mu (f) \text{ to the left-hand side:} \\
\frac{\frac{\partial \mu (f)}{\partial f}}{\mu (f)} =-\frac{1}{2 f} \\
\text{Integrate both sides with respect to }f: \\
\log (\mu (f))=-\frac{\log (f)}{2} \\
\text{Take exponentials of both sides:} \\
\mu (f) =\frac{1}{\sqrt{f}} \\
\text{Multiply both sides of }-2 f(t)+\frac{df(t)}{dt} (-t+2 f(t))=0 \text{ by }\mu (f(t)): \\
-2 \sqrt{f(t)}+\frac{(-t+2 f(t)) \frac{df(t)}{dt}}{\sqrt{f(t)}} =0 \\
\text{Let }P(t,f)=-2 \sqrt{f} \text{ and }Q(t,f)=\frac{2 f-t}{\sqrt{f}}. \\
\text{This is an exact equation, because }\frac{\partial P(t,f)}{\partial f}=-\frac{1}{\sqrt{f}}=\frac{\partial Q(t,f)}{\partial t}. \\
\text{Define }g(t,f) \text{ such that }\frac{\partial g(t,f)}{\partial t}=P(t,f) \text{ and }\frac{\partial g(t,f)}{\partial f}=Q(t,f). \\
\text{Then, the solution will be given by }g(t,f)=c_1\text{, where }c_1\text{ is an arbitrary constant.} \\
\text{Integrate }\frac{\partial g(t,f)}{\partial t} \text{ with respect to }t \text{in order to find }g(t,f): \\
g(t,f)=\int -2 \sqrt{f} \, dt=-2 \sqrt{f} t+h(f)\text{ where }h(f)\text{ is an arbitrary function of }f. \\
\text{Differentiate }g(t,f) \text{ with respect to }f \text{in order to find }h(f): \\
\frac{\partial g(t,f)}{\partial f} =\frac{\partial \text{}}{\partial f}\left(-2 \sqrt{f} t+h(f)\right) =-\frac{t}{\sqrt{f}}+\frac{dh(f)}{df} \\
\text{Substitute into }\frac{\partial g(t,f)}{\partial f}=Q(t,f): \\
-\frac{t}{\sqrt{f}}+\frac{dh(f)}{df}=\frac{2 f-t}{\sqrt{f}} \\
\text{Solve for }\frac{dh(f)}{df}: \\
\frac{dh(f)}{df}=2 \sqrt{f} \\
\text{Integrate }\frac{dh(f)}{df} \text{ with respect to }f: \\
h(f)=\int 2 \sqrt{f} \, df=\frac{4 f^{3/2}}{3} \\
\text{Substitute }h(f) \text{ into }g(t,f): \\
g(t,f)=\frac{4 f^{3/2}}{3}-2 \sqrt{f} t \\
\text{The solution is }g(t,f)=c_1: \\
\frac{4 f^{3/2}}{3}-2 \sqrt{f} t=c_1 \\
\text{Solve for the unknown constants using the initial condition }f(1)=0: \\
\text{Substitute }t=1 \text{ into }\frac{4 f^{3/2}}{3}-2 \sqrt{f} t=c_1: \\
\frac{4 f^{3/2}}{3}-2 \sqrt{f} =c_1 \\
\text{Substitute }f(1)=0 \text{ into }\frac{4 f^{3/2}}{3}-2 \sqrt{f}=c_1: \\
0 =c_1 \\
\text{Solve for }c_1: \\
c_1 =0 \\
\text{Substitute }c_1=0 \text{ into }\frac{4 f^{3/2}}{3}-2 \sqrt{f} t=c_1: \\
\text{Answer:} \\
\frac{4 f^{3/2}}{3}-2 \sqrt{f} t =0 \\
\end{array}

hbghlyj

hbghlyj 发表于 2023-4-16 13:32
两种方法:
\begin{array}l
\text{Answer:}\\
-\frac{1}{3} \log \left(\frac{f(t)}{t}\right)-\frac{2}{3} \log \left(-\frac{2 f(t)}{t}+3\right)=\log (t)+c_1 \\[2ex]
\text{Answer:}\\
\frac{4 f^{3/2}}{3}-2 \sqrt{f} t = 0
\end{array}

两种解法产生不同的答案？第二个等同于 $f=\frac32t$.
它们都不满足$f(1)=0$.
无解

hbghlyj

icesheep 发表于 2014-4-11 05:28
把微分方程看成是 x=x(y) 就能解了

相关帖子: 利用反函数导数公式的三阶非线性ODE
\begin{array}{l}
\text{Solve the linear equation }\frac{dt(f)}{df}+\frac{t(f)}{2 f}=1: \\
\text{Let }\mu (f)=e^{\int \frac{1}{2 f} \, df}=\sqrt{f}. \\
\text{Multiply }\text{both }\text{sides }\text{by }\mu (f): \\
\sqrt{f} \frac{dt(f)}{df}+\frac{t(f)}{2 \sqrt{f}} =\sqrt{f} \\
\text{Substitute }\frac{1}{2 \sqrt{f}}=\frac{d\text{}}{df}\left(\sqrt{f}\right): \\
\sqrt{f} \frac{dt(f)}{df}+\frac{d\text{}}{df}\left(\sqrt{f}\right) t(f)=\sqrt{f} \\
\text{Apply }\text{the }\text{reverse }\text{product }\text{rule }g \frac{dh}{df}+h \frac{dg}{df}=\frac{d\text{}}{df}(g h) \text{ to }\text{the }\text{left-hand }\text{side:} \\
\frac{d\text{}}{df}\left(\sqrt{f} t(f)\right)=\sqrt{f} \\
\text{Integrate }\text{both }\text{sides }\text{with }\text{respect }\text{to }f: \\
\int \frac{d\text{}}{df}\left(\sqrt{f} t(f)\right) \, df =\int \sqrt{f} \, df \\
\text{Evaluate }\text{the }\text{integrals:} \\
\sqrt{f} t(f) =\frac{2 f^{3/2}}{3}+c_1\text{, where }c_1\text{ is an arbitrary constant.} \\
\text{Divide }\text{both }\text{sides }\text{by }\mu (f)=\sqrt{f}: \\
\text{Answer:} \\
t(f) =\frac{2 f}{3}+\frac{c_1}{\sqrt{f}} \\
\end{array}代入$t(0)=1$, 无解