When x∈[0,1], then dxdy+2y=1 ⇒y=21+C1e−2x∵y(0)=0⇒y(x)=21−21e−2x Here, y(1)=21−21e−2=2e2e2−1 When x∈/[0,1], than dxdy+2y=0 ⇒y=c2e−2x∴y(1)=2e2e2−1⇒2e2−1=c2e−2⇒C2=2e2−1∴y(x)(2e2−1)e−2x⇒y(23)=2e3e2−1
Let y=y(x) be the solution of the differential equation dxdy+2y=f(x), where f(x)={1,0,x∈[0,1] otherwise If y(0)=0, then y(23) is
Held on 15 Apr 2018 · Verified 6 Jul 2026.
2e3e2−1
e3e2−1
2e1
2e4e2+1
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