We find the general solution of the equation
dxdyβ+(cotx)y=xcsc(x)
This equation is a linear equation of the form
dxdyβ+P(x)y=Q(x)
where
P(x)=cot(x),Q(x)=xcsc(x)
Integrating factor :
IF(x)=eβ«P(x)dx=eβ«cot(x)dx==eβ«sinxd(sinx)β=elnβ£sinxβ£+C
We are looking for a particular integrating factor, so we take C=0.
Then we get
IF(x)=sin(x)
Multiply both sides of the equation by IF(x)=sin(x)
sin(x)dxdyβ+(sin(x)β
cot(x))y=xcsc(x)β
sin(x)
or
sin(x)dxdyβ+(cos(x))y=x
i.e.
dxdβ(yβ
sinx)=x.
Integrate both sides of this equation we get the general solution
β«dxdβ(yβ
sinx)dx=β«xdxβyβ
sinx=2x2β+C ,
where C - constant defined by the initial conditions.
Now we find C and solve the initial value problem.
Substitute the initial condition
y(2Οβ)=1
into the general solution:
y(2Οβ)β
sin(2Οβ)=2(2Οβ)2β+Cβ1β
sin(2Οβ)=21β(2Οβ)2+C,
or
1=8Ο2β+CβC=1β8Ο2β.
Then we obtain the solution for the initial value problem
yβ
sinx=2x2β+1β8Ο2β
or
y=(2x2β+1β8Ο2β)csc(x)
The solution for the initial value problem is
y=(2x2β+1β8Ο2β)csc(x).
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