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QUESTION


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\item[(a)]
Using the substitution $x=yt$ find the solution of the
differential equation

$$t^2\frac{dx}{dt}=x^2+xt$$

which satisfies the condition $x=1$ when $t=1$.

\item[(b)]
Find the general solution of the second order differential
equation

$$\frac{d^2x}{dt^2}-9x=\cos(4t).$$


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ANSWER

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\item[(a)]
$\ds t^2\frac{dx}{dt}=x^2+xt$\\ Making the substitution $\ds
x=yt,\ \frac{dx}{dt}=y+t\frac{dy}{dt}$, and substituting into the
ODE gives\\ $\ds
t^2\left(y+t\frac{dy}{dt}\right)=t^2y^2+(yt)t=t^2y^2+t^2y$.

Hence $\ds y+t\frac{dy}{dt}=y^2+y$, therefore
$\ds\frac{dy}{dt}=\frac{y^2}{t}$ which is a separable equation.\\
$\ds\int\frac{dy}{y^2}=\int\frac{dt}{t}$ leads to
$\ds-\frac{1}{y}=\ln t+c\Rightarrow y=-\frac{1}{\ln t+c}$\\ When
$t=1,\ y=1$ therefore $\ds1=-\frac{1}{\ln1+c}=-\frac{1}{c},$ so
$c=-1$\\ Therefore $\ds y=-\frac{1}{\ln t-1}=\frac{1}{1-\ln t}$ so
$\ds x=\frac{t}{1-\ln t}$.

\item[(b)]
$\ds\frac{d^2x}{dt^2}-9x=\cos(4t)$\\ To find the complementary
function consider the equation\\ $\ds\frac{d^2x}{dt^2}-9x=0$.\\
This has the auxiliary equation $m^2-9=0,\ m^2=9,\ m=\pm3$, and
therefore the complementary function is $x=Ae^{3t}+Be^{-3t}$\\ To
find a particular integral try $\ds x=C\cos(4t)+D\sin(4t),\\
\frac{dx}{dt}=-4C\sin(4t)+4D\cos(4t),\\
\frac{d^2x}{dt^2}=-16C\cos(4t)-16D\sin(4t)$\\ Substituting this
into the ODE gives\\
$-16C\cos(4t)-16D\sin(4t)-9(C\cos(4t)+D\sin(4t))=\cos(4t)$
\\ i.e. $-25C\cos(4t)-25D\sin(4t)=\cos(4t)$\\ Thus
$\ds-25C=1,-25D=0\rightarrow C=-\frac{1}{25},\ D=0$\\ Hence a
particular integral is $\ds x=-\frac{1}{25}\cos(4t)$ and the
general solution can be written $\ds
x=Ae^{3t}+Be^{-3t}-\frac{1}{25}\cos(4t)$



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