These are all separable first order differential equations.

a)

$\displaystyle{\frac{dy}{dx}=\frac{y}{\var{a1}x} \Rightarrow \int \frac{1}{y}\;dy = \frac{1}{\var{a1}}\int\frac{1}{x}\;dx \Rightarrow \ln(y)=\frac{1}{\var{a1}}\ln(x)+C}$

Exponentiation of both sides then gives $y=Ax^{1/\var{a1}}$ where we have renamed the constant of integration.

To find the particular solution satisfying $y=1$ at $x=2$, we have $\displaystyle{1=A \times 2^{1/\var{a1}} \Rightarrow A = \frac{1}{2^{1/\var{a1}}}}$

Hence the solution is $\displaystyle{y=\left(\frac{x}{2}\right)^{1/\var{a1}}}$

b)

$\displaystyle{\frac{dy}{dx}=-\var{a2}\frac{y}{x} \Rightarrow \int \frac{1}{y}\;dy = -\var{a2}\int\frac{1}{x}\;dx \Rightarrow \ln(y)=-\var{a2}\ln(x)+C}$

Exponentiation of both sides then gives $y=Ax^{-\var{a2}}$ where we have renamed the constant of integration.

The particular solution satisfying $y=1$ at $x=2$, gives $A = 2^{\var{a2}}$

Hence the solution is $\displaystyle{y=\left(\frac{2}{x}\right)^{\var{a2}}}$

c)

$\displaystyle{\frac{dy}{dx}=\var{a3}\frac{x}{y} \Rightarrow \int y\;dy = \var{a3}\int x\;dx \Rightarrow \frac{y^2}{2}=\var{a3}\frac{x^2}{2}+C\Rightarrow y^2=\var{a3}x^2+A}$

The particular solution satisfying $y=1$ at $x=2$, gives $A = \var{1-4*a3}$.

Hence the solution is $\displaystyle{y^2=\simplify[std]{{a3}x^2+{1-4*a3}}}$.

d)

$\displaystyle{\frac{dy}{dx}=-\var{a4}\frac{x}{y} \Rightarrow \int y\;dy = -\var{a4}\int x\;dx \Rightarrow y^2=-\var{a4}x^2+A}$

The particular solution satisfying $y=1$ at $x=2$, gives $A = \var{1+4*a4}$.

Hence the solution is $\displaystyle{y^2=\simplify[std]{{-a4}x^2+{1+4*a4}}}$.

\[\frac{dy}{dx}=\var{a}y^{\var{n}} \Rightarrow y^{\var{-n}}\frac{dy}{dx}=\var{a}\]

Hence on integrating both sides we have $\displaystyle{-\frac{y^{\var{1-n}}}{\var{n-1}}=\var{a}x+ A}$ where $A$ is an arbitrary constant.

It makes calculation a little easier if we replace the arbitrary constant $A$ in this equation by $\displaystyle{-\frac{A}{\var{n-1}}}$ (which is still an arbitrary constant) to get:

\[\begin{eqnarray*} -\frac{y^{\var{1-n}}}{\var{n-1}}&=&\var{a}x- \frac{A}{\var{n-1}}\\ \Rightarrow y^{\var{1-n}}&=& A-\var{a*(n-1)}x\end{eqnarray*} \]
Using the condition $y(0)=\var{b}$ we get $\displaystyle \var{b}=A^{-1/\var{n-1}}\Rightarrow A = \simplify[std]{{b}^{1-n}=1/{b^(n-1)}}$.

This gives \[\displaystyle y^{\var{1-n}}=\simplify[std]{1/{b^(n-1)}-{a*(n-1)}x} \Rightarrow y^{\var{n-1}}=\frac{\var{b^(n-1)}}{1-\var{a*(n-1)*b^(n-1)}x}\]

\[y\frac{dy}{dx}+\var{a}x^2+\var{b}x+\var{c}=0\]
We can separate variables to get:
\[y\frac{dy}{dx}=-\var{a}x^2-\var{b}x-\var{c}\]

and on integrating we get ($A$ is the arbitrary constant of integration):
\[\begin{eqnarray*} \frac{y^2}{2}&=&A - \left(\simplify[std]{{a}/{3}*x^3+{b}/2*x^2+{c}}x\right)\Rightarrow\\ y^2&=&2A- \left(\simplify[std]{{2*a}/{3}*x^3+{b}*x^2+{2*c}}x\right) \end{eqnarray*} \]
Using $y(0)=\var{d}$ gives $2A=\var{d^2}$ and so the solution is:

\[y=\simplify[std]{Sqrt({d ^ 2} - ((({2 * a} / {3}) * (x ^ 3)) + ({b} * (x ^ 2)) + ({(2 * c)} * x)))}\]

We can separate the variables to get

\[\frac{dy}{dx}= \simplify[std]{(1+y^2)/({a}+{b}x)} \Rightarrow \frac{1}{1+y^2}\frac{dy}{dx}=\simplify[std]{{a}+{b}x}\]

On integrating we get:
\[\arctan(y)=\frac{1}{\var{b}}\ln\left(\left|\var{a}+\var{b}x\right|\right)+A \Rightarrow y=\tan\left(\frac{1}{\var{b}}\ln\left(\left|\var{a}+\var{b}x\right|\right)+A\right)\]
To fix the arbitrary constant of integration we use the condition $y(1)=\var{u}$.

As $\arctan(\var{u})=\var{v}$ we see that $\displaystyle{A = \var{v}-\frac{1}{\var{b}}\ln(|\var{a+b}|)}$.

Hence the solution is

\[y=\simplify[std]{tan(({((t * (t -1)) / 2)} * (pi / 3)) + ({((t -1) * (t -2)) / 2} * (pi / 4)) + ((1 / {b}) * ln(abs(({a} + ({b} * x)) / {a + b}))))}\]

$\displaystyle{x\frac{dy}{dx}+\var{a}y=\simplify[std]{{b}x^{n}}}$ is a linear equation of the special type where we can multiply both sides by $\simplify[std]{x^{a-1}}$ to get:
\[\frac{d(x^{\var{a}}y)}{dx}=\simplify[std]{{b}x^{a+n-1}}\]
We can integrate both sides to get:
\[x^{\var{a}}y=\simplify[std]{{b}/{a+n}x^{a+n}+A}\]
to determine $A$ we use the condition $\displaystyle{y(1)=\simplify[std]{{b*(c+1)}/{a+n}}}$ and we see that:
\[A = \simplify[std]{{b*(c+1)}/{a+n} - {b}/{a+n} = {b*c}/{a+n}}\]
and so the solution is:
\[x^{\var{a}}y=\simplify[std]{{b}/{a+n}x^{a+n}+{b*c}/{a+n}} \Rightarrow y=\simplify[std]{{b}/{a+n}*(x^{n}+{c}*x^{-a})}\]

$\displaystyle{\frac{dy}{dx}=\simplify[std]{({a}x+{b}y)/x} \Rightarrow \frac{dy}{dx}=\simplify[std]{{a}+{b}y/x}}$ on dividing out the left hand side of the equation.

Let $y=v\;x$ and we have $\displaystyle{\frac{dy}{dx}=v+x\frac{dv}{dx}}$. Hence on substituting in the equation we get:
\[\begin{eqnarray*} v+x\frac{dv}{dx}&=&\simplify[std]{{a}+{b}*v}\\\Rightarrow x\frac{dv}{dx}&=&\simplify[std]{{a}+{b-1}*v}\\ \Rightarrow \simplify[std]{1/({a}+{b-1}*v)}\frac{dv}{dx}&=&\frac{1}{x} \end{eqnarray*} \]
On integrating we get:
\[\begin{eqnarray*} \simplify[std]{(1 / {b -1}) * ln(Abs({a} + {b -1} * v))} &=& \ln(x) + A_1\\ \Rightarrow\simplify[std]{ ln(Abs({a} + {b -1} * v))}&=&\simplify[std]{{b-1}*ln(x)}+A_2\\ &=&\simplify[std]{ln(x^{b-1})}+A_2 = \simplify[std]{ln(A_3*x^{b-1})} \end{eqnarray*} \]
If we now take exponentials and put $\displaystyle{v=\frac{y}{x}}$ we have:
\[\begin{eqnarray*} \simplify[std]{{a} + {b -1} * y/x}&=&A_3\simplify[std]{x^{b-1}}\\ \Rightarrow\simplify[std]{{b -1} * y/x}&=&A_3\simplify[std]{x^{b-1}}-\var{a}\\ \Rightarrow y&=&\simplify[std]{A_3/{b-1}*x^{b}-{a}/{b-1}*x} \end{eqnarray*} \]
But we know that $\displaystyle{y(1)=\simplify[std]{{c-a}/{b-1}}}$ which gives $\displaystyle{\simplify[std]{{c-a}/{b-1}}=\simplify[std]{A_3/{b-1}-{a}/{b-1}}}$

and we find that $A_3=\var{c}$.

Hence the solution is:
\[y = \simplify[std]{1/{b-1}*({c}x^{b}-{a}x)}\]

\[\frac{dy}{dx}=\simplify[std]{(x^2+y^2+{a}x*y)/({a}*x^2)}\Rightarrow \frac{dy}{dx}=\simplify[std]{1/{a}(1+(y/x)^2)+y/x}\] on dividing out the right hand side of the equation.
Let $y=v\;x$ and we have $\displaystyle{\frac{dy}{dx}=v+x\frac{dv}{dx}}$. Hence on substituting into the equation we get:
\[\begin{eqnarray*} v+x\frac{dv}{dx}&=&\simplify[std]{1/{a}*(1+v^2)+v}\\ \Rightarrow x\frac{dv}{dx}&=&\simplify[std]{1/{a}*(1+v^2)}\\ \Rightarrow \frac{1}{1+v^2}\frac{dv}{dx}&=&\simplify[std]{1/{a}*x} \end{eqnarray*} \]
On integrating we get:
\[\arctan(v)=A_1+\simplify[std]{1/{a}*ln(abs(x))} \Rightarrow v = \tan\left(A_1+\simplify[std]{1/{a}*ln(abs(x))}\right)\]
On putting $\displaystyle{v=\frac{y}{x}}$ we have $\displaystyle{y=x\tan\left(A_1+\simplify[std]{1/{a}*ln(abs(x))}\right)}$

But we know that $y(1)=\var{u}$ hence $\displaystyle{\var{u} = \tan(A_1) \Rightarrow A_1=\var{v}}$

Hence the solution is :
\[y = x\tan\left(\simplify[std]{(({((t * (t -1)) / 2)} * (pi / 3)) + ({((t -1) * (t -2)) / 2} * (pi / 4)) + (1 / {a}) * ln(abs(x)))}\right)\]

On rearranging the equation we get $\displaystyle{\simplify[std]{(x+{b})^{n}*(dx/dt) = {a}}}$ and on integrating we obtain:
$\displaystyle{\simplify[std]{(x+{b})^{n+1}/{n+1}={a}t +A} \Rightarrow x+\var{b}=(A+\var{a*(n+1)}t)^{1/\var{n+1}}}$

Using the condition $x(0)=0$ gives $\displaystyle{A^{1/\var{n+1}}=\var{b} \Rightarrow A=\var{b^(n+1)}}$

Hence the solution is:
\[x(t) = \simplify[std]{({(b ^ (n + 1))} + ({(a * (n + 1))} * t)) ^ (1 / {(n + 1)}) - {b}}\]

If $v(t)$ is the velocity at time $t$ then the acceleration at time $t$ is $\displaystyle{\frac{dv}{dt}}$.

Hence we have the differential equation for the velocity:

\[\frac{dv}{dt}=\frac{\var{a}}{(1+\var{b}t)^{\var{n}}}=\simplify[std]{{a}(1+{b}t)^{-n}}\] where $v(0)=0$ as the object starts from rest.
Integrating this gives: $\displaystyle{v(t) = \simplify[std]{-{a}/{b*(n-1)}*(1+{b}*t)^{-n+1}+A}}$.

Since $v(0)=0$ this gives $\displaystyle{A=\simplify[std]{{a}/{b*(n-1)}}}$ and so the solution is:

$\displaystyle{v(t) = \simplify[std]{{a}/{b*(n-1)}(1-(1+{b}*t)^{-n+1})}}$

It follows that the limiting speed is:

\[\lim_{t \to \infty} v(t) = \simplify[std]{{a}/{b*(n-1)}}m/s\] as

\[(1+\var{b}t)^{\var{-n+1}}=\simplify[std]{1/(1+{b}t)^{n-1}} \rightarrow 0,\;\;t \rightarrow \infty\]