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Chapter 24: Problem 19

Bestimmen Sie alle L枚sungen der Differenzialgleichung $$ \cos x+\sin x+2 \sin x y y^{\prime}=0 $$

Short Answer

Expert verified
Answer: The general solution of the given differential equation is $$y = \pm\sqrt{C_{1} - \ln|- \cos x + \sin x|}$$

Step by step solution

01

Rewrite the given equation

To begin, we will rewrite the given differential equation by isolating \(y^{\prime}\). Doing so, we arrive at: $$ y^{\prime} = -\frac{\cos x +\sin x}{2\sin x y} $$
02

Separate variables

Now that we have the equation in the form \(y^{\prime} = \frac{dy}{dx} = -\frac{\cos x +\sin x}{2\sin x y}\), we will try to separate variables in order to integrate each side individually. Multiply both sides by \(2\sin x y\) and divide by \(\cos x + \sin x\), we have: $$ 2y\sin x\frac{dy}{dx}=-\cos x+\sin x $$ Now let's isolate \(x\)-related terms on the left-hand side and \(y\)-related terms on the right-hand side: $$ \frac{2y\ dy}{-\cos x+\sin x} = \frac{dx}{\sin x} $$
03

Integrate both sides

Now that we have separated the variables, we are ready to integrate both sides. Integrate the left-hand side with respect to \(y\) and the right-hand side with respect to \(x\). $$ \int \frac{2y\ dy}{-\cos x+\sin x} = \int \frac{dx}{\sin x} $$ Let \(u = -\cos x + \sin x\) and notice that \(du = \sin x \ dx\), then $$ \int 2y\ dy = \int\frac{du}{u} $$ Now we integrate: $$ y^2 = C_1 - \ln|u| $$
04

Replace u with the original expression and solve for y

Now, we will substitute back \(u = - \cos x + \sin x\) to obtain the general solution of the differential equation: $$ y^{2} = C_1 - \ln|- \cos x + \sin x| $$ In the final step, we will solve for \(y\): $$ y = \pm\sqrt{C_{1} - \ln|- \cos x + \sin x|} $$ Thus, we have found all the solutions of the given differential equation.

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Most popular questions from this chapter

Man berechne alle partiellen Ableitungen erster und zweiter Ordnung der Funktionen: $$ \begin{aligned} &f(x, y)=x^{2} \mathrm{e}^{y}+\mathrm{e}^{x y} \\ &g(x, y)=\sin ^{2}(x y) \\ &h(x, y)=\mathrm{e}^{\cos x+y^{3}} \end{aligned} $$

Man entwickle die Funktion \(f, \mathbb{R}^{2}\) $$ f(x, y)=y \cdot \ln x+x \mathrm{e}^{y+2} $$ um \(P=\left(\frac{1}{e},-1\right)\) in ein Taylorpolynom zweiter Ordnung.

Man betrachte die Schar aller Strecken von \((0, t)^{\mathrm{T}}\) nach \((1-t, 0)^{\mathrm{T}}\) mit \(t \in[0,1]\) und bestimme die Einh眉llende dieser Strecken.

Man untersuche die Funktion \(f\) $$ f(x, y)= \begin{cases}\frac{x^{3}}{\cos \left(x^{2}+y^{2}\right)-1} & \text { f眉r }(x, y) \neq(0,0) \\ 0 & \text { f眉r }(x, y)=(0,0)\end{cases} $$ auf Stetigkeit im Punkt \((0,0)\).

Untersuchen Sie die beiden Funktionen \(f\) und \(g\), $$ \mathbb{R}^{2} \rightarrow \mathbb{R} $$ $$ f(x)= \begin{cases}\frac{x_{1} x_{2}^{3}}{\left(x_{1}+x_{2}\right)^{2}} & \text { f眉r } x \neq 0 \\ 0 & \text { f眉r } x=0\end{cases} $$ $$ g(x)= \begin{cases}\frac{x_{1}^{3}, x_{2}^{2}}{\left(x_{1}^{2}+x_{2}^{2}\right)^{2}} & \text { f眉r } x \neq 0 \\\ 0 & \text { f眉r } x=0\end{cases} $$ auf Stetigkeit im Ursprung,
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