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# Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:1. y=ex+1 : y′′−y=02. y=x2+2x+C : y′−2x−2=03. y=cosx+C : y′+sinx=04. y=√1+x2 : y′=xy1+x25. y=Ax : xy=y(x≠0)6. y=xsinx : xy=y+x√x2y2(x≠0andx>yorx<y)7. xy=logy+C : y′=y21−xy(xy≠1)8. y−cosy=x : (ysiny+cosy+x)y′=y9. x+y=tan−1y : y2y′+y2+1=010. y=√a2−x2xϵ(−a,a) : x+ydydx=0(y≠0)

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Solution

## 1) Given, y=ex+1; y′′−y=0For ex. y′′−y=0⇒d2ydx2–y=0⇒d2ydx2=y …(1)For this type of equation solunation isy=ex+c C∈RAnd it satisfies the above eqn.So, y=ex+12) y′−2x−2=0⇒dydx=2x+2⇒dy=(2x+2)dxIntegrating on both sides, we get∫dy=∫(2x+2)dxy=x2+2x+c,C∈R3) y′+sinx=0⇒dydx+sin=0⇒dy=sindxIntegrating on both side∫dy=∫sinxdx⇒y=+cosx+c {∫sinxdx=cotx}4) y′=xy1+x2⇒dydx=xy1=x2⇒dyy=xdx1+x2⇒∫dyy=∫xdx1+x2Let x2+1=t2xdx=dt±xdx=dt2 ⇒ln(y)=12 ∫dt(1pt)⇒ln(y)=12[ln(1+2)]+c,c∈R⇒ taking c=0ln(y)=12ln(1+1)=ln(1+1)1/2y=(1+1)1/2y=√1+x25) y=Ax; xy=y(x≠0)xdydx=y⇒dyy=dxx⇒∫dyy=∫dxxln(y)=ln(x)+ln(A)⇒ln(y)=ln(Ax),A∈R⇒y=Ax9) x+y=tan−1y; y2y′+y2+1=0y2y1=−(1+y2)y1=−(1+y2)y2y2y11+y2=(−1)∫y21+y2dy=∫–dx⇒∫1+y2−11+y2dy=−∫dx=∫dy−∫11+y2dy=−∫dx⇒y−tan−1(y)=−x+c⇒x+y=tan−1(y)+c10) y=√a2−x2xϵ(−a,a); x+ydydx=0(y≠0) x+ydydx=0⇒dydx=−xy∫ydy=∫–xdxy22=∫–xdxy2=a−x2y=√a2−x2,xϵ(−a,a)

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