3. `dy/dx+2y=e^x+x , y(0)=1`

The solution of (dy)/(dx) -y=e^(x) , y(0) = 1 , is

The solution of (dy)/(dx)+y=e^(-x), y(0)=0 is

The solution of (dy)/(dx)+y=e^(-x), y(0)=0 "is"

Solve (dy)/(dx)-y=e^(x) ,y(0)=1

The solution of (dy)/(dx)+y=e^(-x),y(0)=0 , is

Solve each of the following initial value problem: (dy)/(dx)+2y=e^(-2x)sin x,y(1)=0

If x^(3) dy + xy dx = x^(2) dy + 2y dx , y (2) = e and x gt 1 , then y(4) is equal to .

For each of the following initial value problems verify that the accompanying functions is a solution. (i) x(dy)/(dx)=1, y(1)=0 => y=logx (ii) (dy)/(dx)=y , y(0)=1 => y=e^x (iii) (d^2y)/(dx^2)+y=0, y(0)=0, y^(prime)(0)=1 => y=sinx (iv) (d^2y)/(dx^2)-(dy)/(dx)=0, y(0)=2, y^(prime)(0)=1 => y=e^x+1 (v) (dy)/(dx)+y=2, y(0)=3 => y=e^(-x)+2