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JEE • MATHEMATICS • DIFFERENTIAL EQUATIONS

Differential equations practice · Medium

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Practice format and chapter quiz

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Showing 451-452 of 452 questions
451 Medium
451 of 452
Application +4 | -1
Let a solution y=y(x)y=y(x) of the differential equation eydy−(2+cos⁡x)dx=0e^y d y-(2+\cos x) d x=0 satisfy y(0)=0y(0)=0 then the value of f(π2)f\left(\frac{\pi}{2}\right) is equal to
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Correct option: B
ln⁡(2+π)\ln (2+\pi)
Option B (ln⁡(2+π)\ln (2+\pi)) is correct because it is the solution to the separable differential equation with the given initial condition.
452 Medium
452 of 452
Application +4 | -1
If y=y(x)y = y(x) is the solution of the differential equation, dydx+2ytan⁡x=sin⁡x, y(π3)=0\frac{{dy}}{{dx}} + 2y\tan x = \sin x,\,y\left( {\frac{\pi }{3}} \right) = 0, then the maximum value of the function y(x)y(x) over RR is equal to:
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Correct option: D
18\frac{1}{8}
Option D (18\frac{1}{8}) is correct because it represents the maximum value of the function y(x) obtained after solving the differential equation and finding the constant of integration.
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