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JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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The differential equation whose solution is y=c1cos⁑ax+c2sin⁑axy=c_1 \cos a x+c_2 \sin a x is

(Where c1,c2c_1, c_2 are arbitrary constants)
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Correct option: B
d2ydx2+a2y=0\frac{d^2 y}{d x^2}+a^2 y=0
Option B (d2ydx2+a2y=0\frac{{d^2 y}}{{d x^2}}+a^2 y=0) is correct because the second derivative of the given solution satisfies this differential equation.

More Questions from DIFFERENTIAL EQUATIONS

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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)
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