Q.

Let the solution y = y(x) of the differential equation dydx–y=1+4 sin x satisfy y(π) = 1. Then y(π2)+10 is equal to __________.          [2024]


Ans.

(7)

Given, dydx–y=1+4 sin x, which is a linear differential equation.

∴  Integrating factor = =e–∫dx=e–x

Hence, solution is ye–x=∫(1+4 sin x)e–xdx

=–e–x+2e–x(–sin x–cos x)+C          ... (i)

y(π) = 1 i.e., at x = π, y = 1

∴  From (i), we get

(1)e–π=–e–π+2e–π(–sin π–cos π)+C

⇒ e–π=–e–π+2e–π+C ⇒ C=0

Hence, y(x) = –1 – 2(sin x + cos x)

Now, y(π2)+10=–1–2(sinπ2+cosπ2)+10

                             = –1 – 2(1) + 10 = 7.