Consider the second order differential equation d2xdt2+adxdt=b,\frac{d^{2}x}{dt^{2}} + a\frac{dx}{dt} = b, where aa and bb are non-zero constants and dxdt≠ba\frac{dx}{dt} \ne \dfrac{b}{a}.

  1. (a)

    By substituting y=dxdty = \dfrac{dx}{dt}, show that the above second order differential equation can be written as

    dydt=b−ay.\frac{dy}{dt} = b - ay.

  2. (b)

    Find yy in terms of tt, and hence find xx in terms of tt.

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Worked solution

1(a)

1. Find an expression for the second derivative in terms of $y$.

Given y=dxdty = \frac{dx}{dt}, we differentiate both sides with respect to tt:
dydt=d2xdt2\frac{dy}{dt} = \frac{d^{2}x}{dt^{2}}

2. Substitute into the original equation and rearrange to show the required form.

Substitute y=dxdty = \frac{dx}{dt} and dydt=d2xdt2\frac{dy}{dt} = \frac{d^{2}x}{dt^{2}} into the given second order differential equation:
dydt+ay=b\frac{dy}{dt} + ay = b
Rearranging to make dydt\frac{dy}{dt} the subject:
dydt=b−ay(shown)\frac{dy}{dt} = b - ay \quad \text{(shown)}

Answer

dydt=b−ay (shown)\frac{dy}{dt} = b - ay \text{ (shown)}

1(b)

1. Separate the variables to solve the differential equation for $y$.

dydt=b−ay\frac{dy}{dt} = b - ay
∫1b−ay dy=∫1 dt\int \frac{1}{b-ay} \, dy = \int 1 \, dt

2. Integrate both sides with respect to their variables.

−1aln⁡∣b−ay∣=t+C-\frac{1}{a} \ln |b - ay| = t + C

3. Make $y$ the subject of the formula.

ln⁡∣b−ay∣=−at−aC\ln |b - ay| = -at - aC
b−ay=±e−at−aCb - ay = \pm e^{-at - aC}
b−ay=Ae−atwhere A=±e−aCb - ay = A e^{-at} \quad \text{where } A = \pm e^{-aC}
y=ba−Aae−aty = \frac{b}{a} - \frac{A}{a} e^{-at}
Let B=−AaB = -\frac{A}{a}.
y=ba+Be−aty = \frac{b}{a} + B e^{-at}

4. Substitute $y = \frac{dx}{dt}$ and set up the integral to find $x$.

dxdt=ba+Be−at\frac{dx}{dt} = \frac{b}{a} + B e^{-at}
x=∫(ba+Be−at) dtx = \int \left( \frac{b}{a} + B e^{-at} \right) \, dt

5. Evaluate the integral to find the general solution for $x$.

x=bat−Bae−at+Dx = \frac{b}{a}t - \frac{B}{a} e^{-at} + D
where BB and DD are arbitrary constants.

Answer

y=ba+Be−aty = \frac{b}{a} + B e^{-at}
x=bat−Bae−at+Dx = \frac{b}{a}t - \frac{B}{a} e^{-at} + D
(where BB and DD are arbitrary constants)

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