dxdy=x7coty−x5excosecydxdy=siny⋅x7coty−sinyx5exsinydxdy−cosy⋅x7=x5−ex let −cosy=tsinydxdy=dxdt
dxdt+x7t=x5−ex I.F. =x7 t. x7=−∫x2exdxcosyx7=x2ex−2∫xexdxcosyx7=x2ex−2xex+2ex+cx=1,y=2π,c=−ecosy=1282e2−e option (3)
If a curve y=y(x) passes through the point (1,2π) and satisfies the differential equation (7x4coty−excosecy)dydx=x5,x≥1, then at x=2, the value of cosy is:
Held on 4 Apr 2025 · Verified 6 Jul 2026.
642e2−e
642e2+e
1282e2−e
1282e2+e
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