Differential Equations 2 Question 4

4. If y=y(x) is the solution of the differential equation dydx=(tanxy)sec2x,xπ2,π2, such that y(0)=0, then yπ4 is equal to

(2019 Main, 10 April I)

(a) 1e2

(b) 12e

(c) 2+1e

(d) e2

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Answer:

Correct Answer: 4. (d)

Solution:

  1. Given differential equation

dydx=(tanxy)sec2xdydx+(sec2x)y=sec2xtanx

which is linear differential equation of the form

dydx+Py=Q

where P=sec2x and Q=sec2xtanx

IF=esec2xdx=etanx

So, solution of given differential equation is

y×IF=(Q×IF)dx+C Let y(etanx)=etanxsec2xtanxdx+Ctanx=tsec2xdx=dtyetanx=ettdt+C=tetetdt+C[ using integration by parts method ]=et(t1)+Cyetanx=etanx(tanx1)+C[t=tanx]y(0)=00=1(01)+CC=1yetanx=etanx(tanx1)+1 Now, at x=π4ye1=e1(11)+1ye1=2e1+1y=e2



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