We have, xdy=(y+x3cosx)dx
⇒xdy=ydx+x3cosxdx
⇒x2xdy−ydx=x2x3cosxdx
⇒dxd(xy)=∫xcosxdx
⇒xy=xsinx−∫1.sinxdx
Therefore, xy=xsinx+cosx+C
At x=π,y=0,
0=−1+C
⇒C=1,x=π,y=0
So, xy=xsinx+cosx+1
⇒y=x2sinx+xcosx+x
Hence, y(2π)=4π2+2π.
Let y=y(x) be the solution of the differential equation xdy=(y+x3cosx)dx with y(π)=0, then y(2π) is equal to:
Held on 25 Jul 2021 · Verified 6 Jul 2026.
4π2+2π
2π2+4π
2π2−4π
4π2−2π
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