Q.

Let y = y(x) be the solution of the differential equation (1–x2)dy=[xy+(x3+2)3(1–x2)]dx, –1 < x < 1, y(0) = 0. If y(12)=mn, m and n are co-prime numbers, then m + n is equal to __________.          [2024]


Ans.

(97)

We have, (1–x2)dy=[xy+(x3+2)3(1–x2)]dx

⇒ dydx=x(1–x2)y+(x3+2)3–3x2(1–x2)

⇒ dydx–x1–x2y=(x3+2)3–3x21–x2

I.F.=e∫–x1–x2dx

Put 1–x2=t ⇒ –xdx=dt2

⇒ I.F. = e∫dt2t=e12ln t=e12ln(1–x2)

=eln(1–x2)12=(1–x2)12=1–x2

∴ y(1–x2)=∫(x3+2)3(1–x2)(1–x2)1–x2dx

⇒ y1–x2=3∫(x3+2)dx

⇒ y1–x2=3[x44+2x]+c

Since, y(0) = 0, so 0=c ⇒ y=31–x2(x44+2x)

Now, y(12)=31–14((12)44+2×12)=2(164+1)

=2(6564)=6532=mn

∴ m = 65, n = 32 ⇒ m + n = 65 +32 = 97.