Second order equations reducible by substitution

Second order equations reducible by substitution Quiz

Spec 5.2 · Edexcel International A Level Further Maths: Quiz

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  1. 1. With x=etx=\mathrm{e}^t, xdydxx\frac{dy}{dx} is equal to:

    • A. etdydt\mathrm{e}^{t}\frac{dy}{dt}
    • B. dydt\frac{dy}{dt}
    • C. e−tdydt\mathrm{e}^{-t}\frac{dy}{dt}
    • D. d2ydt2\frac{d^2y}{dt^2}

  2. 2. With x=etx=\mathrm{e}^t, x2d2ydx2x^2\frac{d^2y}{dx^2} is equal to:

    • A. d2ydt2\frac{d^2y}{dt^2}
    • B. d2ydt2+dydt\frac{d^2y}{dt^2}+\frac{dy}{dt}
    • C. d2ydt2−dydt\frac{d^2y}{dt^2}-\frac{dy}{dt}
    • D. e−2td2ydt2\mathrm{e}^{-2t}\frac{d^2y}{dt^2}

  3. 3. Under x=etx=\mathrm{e}^t, the equation x2y′′+xy′−4y=0x^2y''+xy'-4y=0 becomes:

    • A. y¨+y˙−4y=0\ddot y+\dot y-4y=0
    • B. y¨−y˙−4y=0\ddot y-\dot y-4y=0
    • C. y¨−4y=0\ddot y-4y=0
    • D. y¨+2y˙−4y=0\ddot y+2\dot y-4y=0

  4. 4. The equation y¨+4y˙+4y=0\ddot y+4\dot y+4y=0 has a repeated auxiliary root. With x=etx=\mathrm{e}^t, the solution in terms of xx is:

    • A. y=Ax2+Bx−2y=Ax^2+Bx^{-2}
    • B. y=(A+Bx)e−2xy=(A+Bx)\mathrm{e}^{-2x}
    • C. y=Ax−2+Bln⁡xy=Ax^{-2}+B\ln x
    • D. y=A+Bln⁡xx2y=\frac{A+B\ln x}{x^2}

  5. 5. If y=z e3xy=z\,\mathrm{e}^{3x}, then dydx\frac{dy}{dx} equals:

    • A. (z′+3z)e3x(z'+3z)\mathrm{e}^{3x}
    • B. z′e3xz'\mathrm{e}^{3x}
    • C. (z′+3)e3x(z'+3)\mathrm{e}^{3x}
    • D. 3z′e3x3z'\mathrm{e}^{3x}

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