Question

Find the particular solution of the following differential equation given that at x = 2, y = 1 $x\,{\frac{d y}{d x}}\,+\,2y=x^{2},\,(x\,\neq\,0).$

Find the particular solution of the following differential equation : ${\frac{d y}{d x}}\;=1+x^{2}+y^{2}+x^{2}y^{2},$ given that y = 1, when x = 0.

Find : $\large\textstyle\int\!{\large\frac{x^{2}+x+1}{(x^{2}+1)\left(x+2\right)}}\,d x.$

If A = $\left(\begin{array}{l l }1 & -1 \\ 2 & -1\end{array}\right)$ , B = $\left(\begin{array}{l l }a & 1 \\ b & -1\end{array}\right)$ and (A + B)² = A²+ B², then find the values of a and b.

Find a vector of magnitude 5 units and parallel to the resultant of $\stackrel{\rightarrow}{a}\$ = $2\stackrel{\wedge}{i}+3\stackrel{\wedge}{j} -\stackrel{\wedge}{k}$ and $\stackrel{\rightarrow}{b}\$ = $\stackrel{\wedge}{i}-2\stackrel{\wedge}{j} +\stackrel{\wedge}{k}$.

Solve the differential equation ${\frac{d y}{d x}}\ +\ y\ \cot\ x$= 2 cos x, given that y = 0 when $x={\frac{\pi}{3}}\,.$

Solve the differential equation $(x^{2}\ -\ y x^{2})d y$ + $(y^{2}+x^{2}y^{2})d x=0,$ given that y = 1 when x = 1.