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$\frac{20b}{{\left(4b^3\right)}^3}$

Find the complete solution of the given differential equation
$\frac{dy}{dx}=\frac{3x\sqrt{1+y^2}}{y}$
I know how to solve it if the right side didn't contain either x or y, but I can't think of a way to arrange them. I think I should solve for y, but I'm not sure. Could anyone show me the solution?

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Solve the following equations:
${\log }_2<!--\; \; -->x+8=13$

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Given that x and y satisfy the equation:
${\displaystyle \arctan \left(x\right)+\arctan \left(y\right)+\arctan \left(xy\right)=\frac{11}{12}\pi }$

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Find y which satisfies:
${\displaystyle y^{\prime }=y^a,y\left(a\right)=a-2}$, for ${\displaystyle a\in N}$

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Use a table of integrals with forms involving eu to find the indefinite integral ${\displaystyle \int e^{-4x}\sin 3xdx}$

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Determine whether the Sequence is decreasing or increasing.
I have the sequence $\frac{({10}^n)}{(2n)!}$ and am trying to determine whether the sequence decreases or increases. I feel like the best way to proceed would be to use the squeeze theorem, but am unsure how to apply it to the problem.

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I have
$A=\left(\begin{array}{cc}1& 2\\ 2& 2\\ 2& 1\end{array}\right)$
I have found that $range(A)=\{\left(\begin{array}{c}1\\ 2\\ 2\end{array}\right),\left(\begin{array}{c}2\\ 2\\ 1\end{array}\right)\}$
and that $null(A^T)=\left\{\left(\begin{array}{c}1\\ -<!--\; -\; -->\frac{3}{2}\\ 1\end{array}\right)\right\}$
I have been asked to state the rank and nullity theorem for $A$ and $A^T$. I know the rank and nullity theorem as rank(A) + nullity(A) = n where A is an mxn matrix.
I know that nullity(A) is the number of vectors present in the null space of matrix A, but I'm not too sure how to go on from there.
Can someone help? The solution states dim(null($A^T$)) + dim(range($A$)) = 3

I have a given $2\times <!--\; \times \; -->2$ special linear matrix, for example
$m=\left(\begin{array}{cc}55& 8469\\ 1& 154\end{array}\right)$
and I would like to get the generating form of it from the s and t matrices which are these:
$t=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)s=\left(\begin{array}{cc}0& -<!--\; -\; -->1\\ 1& 0\end{array}\right)$
I don't know if exist an algoritmh for this problem or any proceedings which could help me.

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