Hi im a bit stuck on this question any help would be appreciated
solve the differential equation dy/dx + ycosx/sin x = x, subject to the initial condition y(pi/2) = 3
given
dy/dx+(cosx/sinx)y=x
dy/dx+py=q
which is linear differential equation of first order
here p=cosx/sinx and q=x
Integrating factor= e^int(cosx/sinx)=e^log|sinx| = sinx
solution of the given equation is
y(sinx)=xsinx+c
since y(pi/2)=3
3(sin(pi/2))=pi/2sin(pi/2)+c
3=pi/2+c
3-pi/2=c
hence the solution is
y(sinx)=xsinx+3-pi/2 is the answer
Hi,
given
dy/dx+(cosx/sinx)y=x
dy/dx+py=q
which is linear differential equation of first order
here p=cosx/sinx and q=x
Integrating factor= e^int(cosx/sinx)=e^log|sinx| = sinx
solution of the given equation is
y(sinx)=xsinx+c
since y(pi/2)=3
3(sin(pi/2))=pi/2sin(pi/2)+c
3=pi/2+c
3-pi/2=c
hence the solution is
y(sinx)=xsinx+3-pi/2 is the answer __________________
RK
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