verify that y=C1eaxcosbx+C2eaxsinbx,where C1and C2are arbitary constant, is a solution of the differential equation d2y/dx2-2a(dy/dx)+(a2+b2)y =0 - Maths - Differential Equations

Question

verify that
y=C1eaxcosbx+C2eaxsinbx,where
C1and C2are arbitary constant, is a solution
of the differential equation
d2y/dx2-2a(dy/dx)+(a2+b2)y
=0

Manbar Singh · Accepted Answer

We have,     y = C1eax cos bx + C2eax sin bx⇒y = eax C1
 cos bx + C2 sin bx   
1differentiating 1, with respect to x, we get⇒dydx = eax-b C1 sin bx + b C2 cos bx+C1 cos bx + C2 sin bxaeax⇒dydx =beaxC2 cos bx -C1 sin bx + ay       using 1  
2differentiating 2, with respect to x, we get     d2ydx2 =beax-b C2 sin bx-b C1 cos bx+C2 cos bx -C1 sin bxabeax + adydx⇒d2ydx2 =-b2eaxC1 cos bx+C2 sin bx +abeaxC2 cos bx -C1 sin bx+ adydx⇒d2ydx2 =-b2y + abeaxC2 cos bx -C1 sin bx+ adydx  
3From 2, we getbeaxC2 cos bx -C1 sin bx =dydx - ay       
4from 3 and 4, we get    d2ydx2 =-b2y +adydx - ay +adydx⇒d2ydx2 =-b2y +adydx - a2y +adydx⇒d2ydx2 =2adydx - a2+b2y⇒d2ydx2-2adydx + a2+b2y = 0

verify that y=C1eaxcosbx+C2eaxsinbx,where C1and C2are arbitary constant, is a solution of the differential equation d2y/dx2-2a(dy/dx)+(a2+b2)y =0.

verify that y=C₁e^axcosbx+C₂e^axsinbx,where C₁and C₂are arbitary constant, is a solution of the differential equation d²y/dx²-2a(dy/dx)+(a²+b²)y =0.