Author Topic: TUT 0202 QUIZ5  (Read 607 times)

Jiwen Bi

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TUT 0202 QUIZ5
« on: November 01, 2019, 02:02:53 PM »
$y''+9y=9sec^{2}3t,for\,0< t< \frac{\pi}{6}\\
we\,solve\,homogeneous\,solution\,first\\
y''+9y=0\\
so\,r^{2}+9=0\\
r=\pm 3i\\
homogeneous\,solution\,:y_{c}(t)=c_{1}cos(3t)+c_{2}sin(3t)\\
now\,solve\,the\,general\,solution\\
sub\,c_{1},c_{2}\,to\,u_{1},u_{2}\\
y(t)=u_{1}(t)cos(3t)+u_{2}(t)sin(3t)\\
u_{1}(t)=-\int \frac{y_{2}(t)g(t)}{W(y_{1},y_{2})(t)}dt+c_{1}\\
u_{2}(t)=-\int \frac{y_{1}(t)g(t)}{W(y_{1},y_{2})(t)}dt+c_{2}\\
w(y_{1},y_{2})(t)=3(cos^{2}(3t)+sin^{2}(3t)=3\\
FOR\,y_{1}(t)=cos3t,y_{2}(t)=sin3t,g(t)=9sec^{2}(3t)\\
u_{1}(t)=-\int \frac{sin(3t)9sec^{2}(3t)}{3}dt+c_{1}=-sec(3t)+c_{1}\\
u_{2}(t)=-\int \frac{cos(3t)9sec^{2}(3t)}{3}dt+c_{2}=ln(tan(3t)+sec(3t))
+c_{2}\\
y(t)=homogeneous eend ous+particular\\
y(t)=c_{1}cos(3t)+c_{2}sin(3t)+(ln(tan(3t)+sec(3t))+c_{2})sin(3t)-1
w(y_{1},y_{2})(t)=3(cos^{2}(3t)+sin^{2}(3t)=3\\
u_{1}(t)=-\int \frac{sin(3t)9sec^{2}(3t)}{3}dt+c_{1}=-sec(3t)+c_{1}\\
u_{2}(t)=-\int \frac{cos(3t)9sec^{2}(3t)}{3}dt+c_{2}=ln(tan(3t)+sec(3t))
+c_{2}\\
y(t)=homogeneous eend ous+particular\\
y(t)=c_{1}cos(3t)+c_{2}sin(3t)+(ln(tan(3t)+sec(3t))+c_{2})sin(3t)-1
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