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Question

Solve each of the following initial value problems:
(i) y'+y=ex, y0=12

(ii) xdydx-y=log x, y1=0

(iii) dydx+2y=e-2x sin x, y0=0

(iv) xdydx-y=x+1e-x, y1=0

(v) 1+y2 dx+x-e-tan-1y dx=0, y0=0

(vi) dydx+y tan x=2x+x2 tan x, y0=1

(vii) xdydx+y=x cos x+sin x, yπ2=1

(viii) dydx+y cot x=4x cosec x, yπ2=0

(ix) dydx+2y tan x=sin x; y=0 when x=π3

(x) dydx-3y cot x=sin 2x; y=2 when x=π2

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Solution

i We have,y'+y=exdydx+y=ex .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=1 and Q=ex I.F.=eP dx =e1 dx = exMultiplying both sides of 1 by I.F.=ex, we getex dydx+y=exexexdydx+exy=e2xIntegrating both sides with respect to x, we gety ex=e2x dx+Cy ex=e2x2+C .....2Now, y0=12 12 e0=e02+CC=0Putting the value of C in 2, we getyex=e2x2ex=ex2Hence, y=ex2 is the required solution.


ii We have,xdydx-y=log xdydx-yx=log xx .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-1x and Q=log xx I.F.=eP dx =e-1x dx =e-log x =1xMultiplying both sides of 1 by I.F.=1x, we get1x dydx-1xy=1x×logxx1xdydx-1x2y=logxx2Integrating both sides with respect to x, we gety1x=1x2II×logxI dx+Cyx=log x1x2dx-ddxlog x1x2dxdx+Cyx=-log xx+1x2dx+Cyx=-log xx-1x+Cy=-log x-1+Cx .....2Now, y1=0 0=-0-1+C1C=1Putting the value of C in 2, we gety=-log x-1+xy=x-1-log xHence, y=x-1-log x is the required solution.


iii We have, dydx+2y=e-2xsin x .....1Clearly, it is a linear differential equation of the formdydx+Py=Qwhere P=2 and Q=e-2xsin x I.F.=eP dx =e2 dx = e2xMultiplying both sides of 1 by I.F.=e2x, we gete2x dydx+2y=e2xe-2xsin xe2x dydx+2y=sin xIntegrating both sides with respect to x, we getye2x=sin x dx+Cye2x=-cos x+C .....2Now,y0=0 0×e0=-cos 0+CC=1Putting the value of C in 2, we getye2x=-cos x+1ye2x=1-cos xHence, ye2x=1-cos x is the required solution.


iv We have,xdydx-y=x+1e-xdydx-1xy=x+1xe-x .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-1x and Q=x+1xe-x I.F.=eP dx =e-1x dx =e-log x =1xMultiplying both sides of 1 by I.F.=1x, we get1x dydx-1xy=1xx+1xe-x 1x dydx-1xy=x+1x2e-xIntegrating both sides with respect to x, we get1xy=1x+1x2e-x dx+CPutting 1xe-x=t-1xe-x-1x2e-xdx=dt1x+1x2e-x dx=-dt1xy=-dt+Cyx=-t+Cyx=-e-xx+Cy=-e-x+Cx .....2Now, y1=0 0=-e-1+CC=e-1Putting the value of C in 2, we gety=-e-x+xe-1y=xe-1-e-xHence, y=xe-1-e-x is the required solution.


v We have,1+y2dx+x-e-tan-1ydy=0x-e-tan-1ydydx=-1+y21+y2dxdy=-x-e-tan-1ydxdy+x1+y2=e-tan-1 y1+y2 .....1Clearly, it is a linear differential equation of the form dxdy+Px=Qwhere P=11+y2 and Q=e-tan-1 y1+y2 I.F.=eP dy =e11+y2 dy =etan-1yMultiplying both sides of 1 by I.F.=etan-1y, we getetan-1y dxdy+x1+y2=etan-1ye-tan-1 y1+y2etan-1y dxdy+x1+y2=11+y2Integrating both sides with respect to y, we getetan-1yx=11+y2 dy+Cxetan-1y=tan-1y+C .....2Now, y0=0 0×e0=0+CC=0Putting the value of C in 2, we getxetan-1y=tan-1y+0xetan-1y=tan-1yHence, xetan-1y=tan-1y is the required solution.


vi We have,dydx+y tan x=2x+x2tan x .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=tan x and Q=x2cot x+2x I.F.=eP dx =etan x dx = elogsec x=sec xMultiplying both sides of 1 by I.F.=sec x, we getsec xdydx+ytan x=sec xx2tan x+2xsec xdydx+ytan x=x2tan x sec x+2x sec xIntegrating both sides with respect to x, we getysec x=x2tan xsec x dx+2xII sec xI dx +Cy sec x=x2tan x sec x dx+2sec xx dx-2ddxsec xx dxdx+Cy sec x=x2tan x sec x dx+x2sec x-x2tan x sec x dx+Cy sec x=x2sec x+Cy=x2+Ccos x .....2Now,y0=1 1=0+Ccos 0C=1Putting the value of C in 2, we gety=x2+cos xHence, y=x2+cos x is the required solution.


vii We have,xdydx+y=x cos x+sin xdydx+1xy=cos x+sin xx .....1Clearly, it is a linear differential equation of the formdydx+Py=Qwhere P=tan x and Q=x2cot x+2x I.F.=eP dx =e1xdx = elog x=xMultiplying both sides of (1) by I.F.=x, we getxdydx+1xy=xcos x+sin xxxdydx+1xy=x cos x+sin xIntegrating both sides with respect to x, we getxy=x cos x dx+sin x dx+Cxy=x sin x-1sin xdx-cos x+Cxy=x sin x+cos x-cos x+Cxy=x sin x+C .....2Now, yπ2=1 1×π2=π2sinπ2+CC=0Putting the value of C in 2, we getxy=x sin xy=sin xHence, y=sin x is the required solution.


viii We have, dydx+y cot x=4x cosec x .....1Clearly, it is a linear differential equation of the formdydx+Py=Qwhere P=cot x and Q=4x cosec x I.F.=eP dx =ecot x dx = elogsin x=sin xMultiplying both sides of 1 by I.F.=sin x, we getsin xdydx+y cot x=sin x4x cosec xsin xdydx+y cot x=4xIntegrating both sides with respect to x, we gety sin x=4x dx+Cy sin x=2x2+C .....2Now,yπ2=0 0×sinπ2=2π22+CC=-π22Putting the value of C in 2, we gety sin x=2x2-π22Hence, y sin x=2x2-π22 is the required solution.


ix We have,dydx+2y tan x=sin x .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=2tan x and Q=sin x I.F.=eP dx =e2tan x dx = e2logsec x=sec2xMultiplying both sides of (1) by I.F.=sec2 x, we getsec2 x dydx+2y tan x=sec2 x×sin xsec2 x dydx+2y tan x=tan x sec xIntegrating both sides with respect to x, we gety sec2 x=tan x sec x dx+Cy sec2 x=sec x+C .....2Now, yπ3=0 0secπ32=secπ3+CC=-2Putting the value of C in 2, we gety sec2 x=sec x-2y=cos x-2cos2 xHence, y=cos x-2cos2 x is the required solution.


x We have, dydx-3y cot x=sin 2x .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-3cot x and Q=sin 2x I.F.=eP dx =e-3cot x dx = e-3logsin x=cosec3xMultiplying both sides of 1 by I.F.=cosec3x, we getcosec3xdydx-3y cot x=sin 2xcosec3xcosec3xdydx-3y cot x=2cot x cosec xIntegrating both sides with respect to x, we gety cosec3x=2cot x cosec x dx+C ycosec3x=-2cosec x+Cy=-2sin2x+Csin3x .....2Now, yπ2=2 2=-2sin2π2+Csin3 π2C=4Putting the value of C in 2, we gety=-2sin2x+4sin3xy=4sin3x-2sin2xHence, y=4sin3x-2sin2x is the required solution.

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