Równanie różniczkowe.

[opop1995]

Rozwiąż zagadnienie początkowe:

\(tgtdy+( \frac{y(t)}{cos^2t} + \frac{1}{t^2-1})dt=0\)
\(y( \frac{ \pi }{4})=ln \sqrt{ \frac{4+ \pi }{4- \pi } }\)

[octahedron]

\(\displaystyle{
\frac{d}{dt}\tg t=\frac{1}{\cos^2t}=\frac{d}{dy}\left(\frac{y}{\cos^2t}+\frac{1}{t^2-1}\right)\quad\Rightarrow\text{ równanie zupełne}\\
F(x,y)=\int\tg t\,dy=y\tg t+C(t)\\
\frac{d}{dt}F=\frac{y}{\cos^2t}+C'(t)=\frac{y}{\cos^2t}+\frac{1}{t^2-1}\\
C(t)=\int\frac{1}{t^2-1}\,dt=\frac{1}{2}\int\frac{1}{t-1}-\frac{1}{t+1}\,dt=\frac{1}{2}(\ln|t-1|-\ln|t+1|)+D=\frac{1}{2}\ln\left|\frac{t-1}{t+1}\right|+D\\
F(x,y)=y\tg t+\frac{1}{2}\ln\left|\frac{t-1}{t+1}\right|+D=0\\
y=\frac{1}{2}\ctg t\ln\left|\frac{t+1}{t-1}\right|+D\\
y\left(\frac{\pi}{4}\right)=\frac{1}{2}\ln\frac{4+\pi}{4-\pi}+D\quad\Rightarrow\quad D=0\\
}\)

[opop1995]

A mógłby ktoś zrobić ten przykład metodą czynnika całkującego?

[opop1995]

Mógłby ktoś??

[octahedron]

Przecież nie jest potrzebny.

[Robakks]

To równanie jest liniowe

\(\tan{t}\frac{\mbox{d}y}{\mbox{d}t}+\frac{y}{\cos^{2}{t}}=-\frac{1}{t^2-1}\\
\tan{t}\frac{\mbox{d}y}{\mbox{d}t}+\frac{y}{\cos^{2}{t}}=0\\
\frac{\sin{t}}{\cos{t}}\frac{\mbox{d}y}{\mbox{d}t}=-\frac{y}{\cos^{2}{t}}\\
\sin{t}\frac{\mbox{d}y}{\mbox{d}t}=-\frac{y}{\cos{t}}\\
\frac{\mbox{d}y}{y}=-\frac{1}{\cos{t}\sin{t}}\mbox{d}t\\
\frac{\mbox{d}y}{y}=-\frac{\cos^{2}{t}+\sin^{2}{t}}{\cos{t}\sin{t}}\mbox{d}t\\
\frac{\mbox{d}y}{y}=- \left(\frac{\cos{t}}{\sin{t}}-\frac{ \left(-\sin{t} \right) }{\cos{t}} \right)\mbox{d}y\\
\ln{ \left| y\right| }=-\ln{ \left|\sin{t} \right| }+\ln{ \left| \cos{t}\right| }+C\\
\ln{ \left| y\right| }=\ln{ \left|\frac{\cos{t}}{\sin{t}}\right| }+C\\
y=C\cdot \frac{\cos{t}}{\sin{t}}\\
y \left( t\right)=C \left(t \right) \cdot \frac{\cos{t}}{\sin{t}} \\
\tan{t} \left( C '\left(t \right)\cot{t}-C \left( t\right)\frac{1}{\sin^{2}{t}} \right) +C \left( t\right)\frac{1}{\sin{t}\cos{t}}=-\frac{1}{t^2-1}\\
C '\left(t \right)-C \left( t\right)\frac{1}{\sin{t}\cos{t}}+C \left( t\right)\frac{1}{\sin{t}\cos{t}}=-\frac{1}{t^2-1}\\
C '\left(t \right)=\frac{1}{2}\frac{\left(t-1\right)-\left(t+1\right)}{t^2-1}\\
C '\left(t \right)=\frac{1}{2}\frac{1}{t+1}-\frac{1}{2}\frac{1}{t-1}\\
C \left(t \right)=\frac{1}{2}\ln{ \left|\frac{t+1}{t-1} \right| }+C_{1}\\
y=\frac{1}{2}\ln{ \left|\frac{t+1}{t-1} \right| }\cdot \frac{\cos{t}}{\sin{t}}+C_{1}\frac{\cos{t}}{\sin{t}}\\\)

\(y( \frac{ \pi }{4})=\ln \sqrt{ \frac{4+ \pi }{4- \pi } }\\
y( \frac{ \pi }{4})=\frac{1}{2}\ln{ \left|\frac{\pi+4}{\pi-4} \right| }+C_{1}\\
\ln \sqrt{ \frac{4+ \pi }{4- \pi } }=\frac{1}{2}\ln{ \left|\frac{\pi+4}{\pi-4} \right| }+C_{1}\\
C_{1}=\ln \sqrt{ \frac{4+ \pi }{4- \pi } }-\frac{1}{2}\ln{ \left|\frac{\pi+4}{\pi-4} \right| }\\\)

[octahedron]

\(\displaystyle\frac{1}{2}\ln\left|\frac{\pi+4}{\pi-4}\right|=\frac{1}{2}\ln\frac{4+\pi}{4-\pi}=\ln\sqrt{\frac{4+\pi}{4-\pi}}\quad\Rightarrow\quad C_1=0\)

[Robakks]

Z tym czynnikiem całkującym to chodziło chyba o to aby rozwiązywać to równanie jak równanie liniowe