Skip to main content
Back to DIFFERENTIAL EQUATIONS

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

1 Easy
1 of 1
Analyzing +4 | -1
If c1,c2,c3,c4,c5{c_1},{c_2},{c_3},{c_4},{c_5} and c6{c_6} are constants, then the order of the differential equation whose general solution is given by
y=c1cos⁑(x+c2)+c3sin⁑(x+c4)+c5ex+c6y = {c_1}\cos (x + {c_2}) + {c_3}\sin (x + {c_4}) + {c_5}{e^x} + {c_6} is
View answer and explanation
Correct option: C
4
Option C (4) is correct because the given expression can be simplified to contain four independent arbitrary constants, thus the order of the differential equation is 4.

More Questions from DIFFERENTIAL EQUATIONS

Practice All Chapter MCQs β†’
Let y=y(x)y=y(x) be the solution of the differential equation (3y2βˆ’5x2)ydx+2x(x2βˆ’y2)dy=0\left(3 y^{2}-5 x^{2}\right) y d x+2 x\left(x^{2}-y^{2}\right) d y=0 such that y(1)=1y(1)=1. Then ∣(y(2))3βˆ’12y(2)∣\left|(y(2))^{3}-12 y(2)\right| is equal to : The degree and order of the differential equation whose solution is a parabola whose axis is xx-axis, are If x2+y2=1{x^2} + {y^2} = 1, then(yβ€²=dydx,yβ€²β€²=d2ydx2)\left( {y' = \frac{{dy}}{{dx}},y'' = \frac{{{d^2}y}}{{d{x^2}}}} \right) If x=Acos⁑4t+Bsin⁑4t,x = A\cos 4t + B\sin 4t, then d2xdt2\frac{{{d^2}x}}{{d{t^2}}} is equal to Let y=y1(x)y=y_{1}(x) and y=y2(x)y=y_{2}(x) be the solution curves of the differential equation dydx=y+7\frac{d y}{d x}=y+7 with initial conditions y1(0)=0y_{1}(0)=0 and y2(0)=1y_{2}(0)=1 respectively. Then the curves y=y1(x)y=y_{1}(x) and y=y2(x)y=y_{2}(x) intersect at Let y=y(t)y=y(t) be a solution of the differential equation dydt+Ξ±y=Ξ³eβˆ’Ξ²t\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}
where, Ξ±>0,Ξ²>0\alpha>0, \beta>0 and Ξ³>0\gamma>0. Then lim⁑xβ†’βˆžy(t)\lim \limits_{{x \rightarrow \infty}} y(t)
Type at least 2 characters or jump directly:
πŸ”“

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.