/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Berechnen Sie die L枚sungen der ... [FREE SOLUTION] | 91影视

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Chapter 20: Problem 7

Berechnen Sie die L枚sungen der folgenden Anfangswertprobleme: (a) \(u^{\prime}(x)=\frac{x}{3 \sqrt{1+x^{2}}(u(x))^{2}}, \quad x>0\) \(u(0)=3\) (b) \(u^{\prime}(x)=-\frac{1}{2 x} \frac{(u(x))^{2}-6 u(x)+5}{u(x)-3}, \quad x>1\) \(u(1)=2\)

Short Answer

Expert verified
The solution to the given IVP is: \(u(x) = \sqrt[3]{27 + 2\ln{({1+x^2})}}\) b) Can the IVP u'(x)=(-1/2 x)((u(x))^2-6u(x)+5)/(u(x)-3) with u(1)=2 be directly solved for u(x) in an elementary function? No, the IVP u'(x)=(-1/2 x)((u(x))^2-6u(x)+5)/(u(x)-3) with u(1)=2 cannot be directly solved for u(x) in an elementary function. However, numerical methods can be employed to obtain an approximate solution for specific values of x.

Step by step solution

01

Separate variables

Let's rewrite the equation into separate variables form: \(\frac{du}{dx} = \frac{x}{3 \sqrt{1+x^{2}}(u(x))^{2}}\) Separate the variables: \((u(x))^2 du = \frac{x}{3\sqrt{1+x^2}} dx\)
02

Integrate both sides

Now, integrate both sides with respect to their corresponding variables: \(\int (u(x))^2 du = \int \frac{x}{3 \sqrt{1+x^2}} dx\)
03

Evaluate the integrals and solve for u(x)

Solve the integrals: \((\frac{1}{3}u^3)|_{u(0)}^{u(x)} = \frac{1}{6}\ln(1+x^2)|_{0}^{x}\) With the initial condition, u(0) = 3, plug it in the limit: \(\frac{1}{3}(u^3 - 27) = \frac{1}{6}\ln{(1+x^2)}\) Now, solve for u(x): \(u^3(x) = 27 + 2\ln{(1+x^2)}\) \(u(x) = \sqrt[3]{27 + 2\ln{(1+x^2)}}\) The solution to the given IVP is: \(u(x) = \sqrt[3]{27 + 2\ln{({1+x^2})}}\) #b) u'(x)=(-1/2 x)((u(x))^2-6u(x)+5)/(u(x)-3) with u(1)=2#
04

Separate variables

Rewrite the equation into separate variables form: \(\frac{du}{dx} = -\frac{1}{2x} \frac{(u(x))^{2}-6u(x)+5}{u(x)-3}\) Separate the variables: \((u(x)-3)du = -\frac{1}{2x}(u^2(x)-6u(x)+5) dx\)
05

Integrate both sides

Now, integrate both sides with respect to their corresponding variables: \(\int (u(x)-3) du = -\frac{1}{2} \int \frac{(u^2(x)-6u(x)+5)}{x} dx\)
06

Evaluate the integrals and solve for u(x)

Solve the integrals: \(\frac{1}{2}u^2(x)-3u(x)|_{u(1)}^{u(x)}=-\frac{1}{2}(u^2(x)-6u(x)+5)\ln{|x|}|_{1}^{x}\) With the initial condition, u(1) = 2, plug it in the limit: \(\frac{1}{2}(u^2(x)-8u(x)+12)=-\frac{1}{2}(u^2(x)-6u(x)+5)\ln{(x)}\) Now, solve for u(x): \((u^2(x)-8u(x)+12)=(u^2(x)-6u(x)+5)\ln{(x)}\) It is a first-order linear inhomogeneous differential equation with a logarithmic term. This equation cannot be directly solved for u(x) in an elementary function. However, numerical methods can be employed to obtain an approximate solution for specific values of x.

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Most popular questions from this chapter

$$ \text { det }: \mathbb{R}^{n \times n}=\left(\mathbb{R}^{n}\right)^{n} \rightarrow \mathbb{R},\left(a_{1}, \ldots, \boldsymbol{a}_{n}\right) \mapsto \operatorname{det}\left(a_{1}, \ldots, a_{n}\right) $$ Zeigen Sie: F眉r das Differenzial gilt $$ \begin{aligned} &\mathrm{d}\left(\operatorname{det}\left(a_{1}, \ldots, a_{n}\right)\right)\left(h_{1}, \ldots, h_{n}\right) \\ &\quad=\sum_{j=1}^{n} \operatorname{det}\left(a_{1}, \ldots, \boldsymbol{a}_{j-1}, \boldsymbol{h}_{j}, \boldsymbol{a}_{j+1}, \ldots, \boldsymbol{a}_{n}\right) \end{aligned} $$

Berechnen Sie die ersten drei sukzessiven Iterationen zu dem Anfangswertproblem $$ u^{\prime}(x)=x-(u(x))^{2}, \quad x \in \mathbb{R}, \quad u(0)=1 $$

Sei \(D \subseteq \mathbb{R}^{n}\) offen (und \(\left.\neq \emptyset\right)\) und \(f: D \rightarrow \mathbb{R}\) in \(a\) total differenzierbar. (a) Warum existieren dann alle Richtungsableitungen \(\partial_{v} f(\boldsymbol{a}) ?\) (b) Welcher Zusammenhang besteht zwischen \(\partial_{v} f(\boldsymbol{a})\) und \(\operatorname{grad} f(\boldsymbol{a}) ?\) (c) Falls grad \(f(a) \neq 0\) gilt, warum ist dann \(\|\operatorname{grad} f(\boldsymbol{a})\|_{2}=\max \left\\{\partial_{v} f(\boldsymbol{a}) \mid\|\boldsymbol{v}\|_{2}=1\right\\}\) (d) Ist \(f\) in allen Punkten \(x \in D\) total differenzierbar und $$ \alpha: M \rightarrow D, t \mapsto\left(\alpha_{1}(t), \ldots, \alpha_{n}(t)\right)^{\top} $$ \((M \subseteq \mathbb{R}\) ein Intervall), und gibt es ein \(c \in \mathbb{R}\) mit \(f(\boldsymbol{\alpha}(t))=c\), warum gilt dann \(\operatorname{grad} f(\boldsymbol{\alpha}(t))\) und \(\dot{\alpha}(t)=\left(\dot{\alpha}_{1}(t), \ldots \dot{\alpha}_{n}(t)\right)^{\top}\) orthogonal?

$$ f(x, y)=\left\\{\begin{aligned} x y \frac{x^{2}-y^{2}}{x^{2}+y^{2}} & \text { f眉r }(x, y)^{\top} \neq(0,0)^{\top} \\ 0 & \text { f眉r }(x, y)^{\top}=(0,0)^{\top} \end{aligned}\right. $$ Zeigen Sie: (a) \(f\) ist in \(\mathbb{R}^{2}\) stetig partiell differenzierbar. (b) \(f\) ist in \(\mathbb{R}^{2} \backslash\\{0\\}\) beliebig oft stetig partiell differenzierbar. (c) \(f\) ist in \((0,0)^{\top}\) zweimal partiell differenzierbar, aber es gilt \(\partial_{2} \partial_{1} f(0,0) \neq \partial_{1} \partial_{2} f(0,0)\). (d) Ist dies ein Widerspruch zum Vertauschungssatz von Schwarz?

Bestimmen Sie die allgemeine L枚sung der linearen Differenzialgleichung erster Ordnung: $$ u^{\prime}(x)+\cos (x) u(x)=\frac{1}{2} \sin (2 x), \quad x \in(0, \pi) $$
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