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GATE [ME, 2017, 2 marks] Question : Consider the differential equation $3y''(x)+27y(x)=0$ with initial conditions $y(0)=0$ and $y'(0)=2000$. The value of $y$ at $x=1$ is____. Solution : Given $3y''(x)+27y(x)=0$ $\implies (3D^2+27)y=0$ The auxiliary equation is $3D^2+27=0$ $\implies D=\pm 3i$ General solution is $y=C_1\cos 3x+C_2\sin 3x\quad\quad\quad\ldots\text{(1)}$ differentiating equation $(1)$ $y'=-3C_1\sin 3x+3C_2\cos 3x$ $y(0)=0$ $\implies C_1=0$ $y'(0)=2000$ $3C_2=2000$ $\implies C_2=\dfrac{200}{3}$ $\therefore$ the particular solution is $y=\dfrac{2000\sin 3x}{3}$ At $x=1$, $y=\dfrac{2000\sin 3}{3}$ $y=94.08$ CORRECT ANSWER : 94.08

GATE [EE, 2010, 2 marks] Question : For the differential equation $\displaystyle\frac{d^2x}{dt^2}+6\frac{dx}{dt}+8x=0$ with initial conditions $x(0)=1$ and $\displaystyle\frac{dx}{dt}\Bigg|_{t=0}=0$, the solution is (a) $x(t)=2e^{-6t}-e^{-2t}$ (b) $x(t)=2e^{-2t}-e^{-4t}$ (c) $x(t)=-e^{-6t}+2e^{-4t}$ (d) $x(t)=e^{-2t}+2e^{-4t}$ Solution : Given $\dfrac{d^2x}{dt^2}+6\dfrac{dx}{dt}+8x=0$ Let $\dfrac{d}{dt}=D$ so the equation becomes $\implies (D^2+6D+8)x=0$ the auxiliary equation is $\implies D^2+6D+8=0$ $\implies (D+2)(D+4)=0$ $\implies D=-2,-4$ the solution is $\implies x=c_1e^{-2t}+c_2e^{-4t}\quad\quad\quad\quad\ldots(1)$ differentiting equation $(1)$ $\implies\dfrac{dx}{dt}=-2c_1e^{-2t}-4c_2e^{-4t}\quad\quad\quad\quad\ldots(2)$ since $x(0)=1$ and $\dfrac{dx}{dt}\Bigg|_{t=0}=0$ putting values in equations $(1)$ and $(2)$ respectively $1=c_1+c_2\quad\quad\quad\quad\ldots(3)$ $0=-2c_1-4c_2\quad\quad\quad\quad\ldots(4)$ solving $3$ and $(4)$ $\implies c_1=2, c_2...