(dy)/(dx)+2y=e^(-2x)sin x,y(0)=0

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

Solve the following initial value problem: (dy)/(dx)+2y=e^(-2x)sinx ,\ y\ (0)=0

(dy)/(dx) + 2y = e ^(-2x) sinx , given that y=0 when x = 0

If (dy)/(dx)= y sin 2x, y(0)=1 , then solution is

(d^(2)y)/(dx^(2))=x^(2)sin x, givenatx =0 if y=0,(dy)/(dx)=1

The solution of (dy)/(dx)=e^(x)(sin^(2)x+sin2x)/(y(2log y+1)) is

If y=e^tanx then prove that: cos^2x(d^2y)/(dx^2)-(1+sin2x)(dy)/dx=0

If y=e^tanx then prove that: cos^2x(d^2y)/(dx^2)-(1+sin2x)(dy)/dx=0

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