{F} 2. Find the particular solution of the differential equation 1 − cos(2t) dx/dt = x sin(2t) given that x(π/2) = 2.
Let us find the particular solution of the differential equation "(1 \u2212 \\cos(2t))\\frac{ dx}{dt} = x \\sin(2t)" given that "x(\\frac{\u03c0}2) = 2."
This equation is equivalent to "\\frac{ dx}{x} = \\frac{\\sin(2t)dt}{1 \u2212 \\cos(2t)}", and hence "\\int\\frac{ dx}{x} = \\int\\frac{\\sin(2t)dt}{1 \u2212 \\cos(2t)}" .
It follows that
"\\ln|x|= \\frac{1}2\\int\\frac{d(1 \u2212 \\cos(2t))}{1 \u2212 \\cos(2t)}=\\frac{1}2\\ln|1 \u2212 \\cos(2t)|+\\ln|C|=\\ln|C\\sqrt{1 \u2212 \\cos(2t)}|."
We conclude that "x(t)=C\\sqrt{1 \u2212 \\cos(2t)}" is the general solution.
If "x(\\frac{\u03c0}2) = 2" then "2=C\\sqrt{1-\\cos\\pi}=C\\sqrt{2}." It follows that "C=\\sqrt{2}," and thus the particular solution is
"x(t)=\\sqrt{2}\\sqrt{1 \u2212 \\cos(2t)}=\\sqrt{2 \u2212 2\\cos(2t)}."
Comments
Leave a comment