Question

y=e^ax sin bx prove that y2-2ay1+(a^2+b^2)y=0

Answer 1

y=[e^(ax)][sin(bx)]
Let's differentate :
y'=[e^(ax)]'[sin(bx)]+[e^(ax)][sin(bx)]'
y'=a[e^(ax)][sin(bx)]+[e^(ax)]b[cos(bx)]
Let's differentate 2nd time :
y''={ a[e^(ax)][sin(bx)]+[e^(ax)]b[cos(bx)] } '
y''={ a[e^(ax)][sin(bx)]}' + { [e^(ax)]b[cos(bx)] }'
y''= (a^2)[e^(ax)][sin(bx)]+a[e^(ax)]b[cos(bx)] + a[e^(ax)]b[cos(bx)]-[e^(ax)](b^2)[sin(bx)].

y''=2ay'-(a^2+b^2)y .

Answer 2

Well line by line I end up with something like this

\[ a^2\sin bx + ab \cos bx -b^2 \sin bx + ab \cos bx \\-2 a^2 \sin bx -2ab \cos bx \\ \underline{a^2 \sin bx + b^2 \sin bx }
\\=0 \]