$y'+y=x^2$

$\frac{\left(1-\cos\left(x\right)\right)\left(1+\cos\left(x\right)\right)}{\cos\left(x\right)}$

$-3x+y-4\cdot3x^2$

$x ^ { 2 } + 6 x y + 9 y ^ { 2 } = ( x + 3 y ) ^ { 2 }$

$\frac{4x+6}{2x+3}$

$3a+1;\:a=2$

$\int\left(\frac{2x-3}{\left(x-3\right)^3}\right)dx$