Solve the initial value problem:
π₯^2π¦" + π₯π¦β² + 9π¦ = 0; π¦(1) = 2, π¦β²(1) = 0
Find the general solution of the equation:
π₯^2π¦" β 7π₯π¦β² + 12π¦ = 0
Find the solution of partial differentail equation ,whether it is linear,non linear or quazi linear .Uxx+xUy=y
Find the integrating factor and solve the following equations:
( π¦ β π₯^2) ππ₯ +( π₯^2sin π¦ β π₯ )ππ¦ = 0
solve (1+tΒ²)y' +4ty=(1+tΒ²)^-2; y(0)=1
y''-y=3x^2e^x
solve the ivp cos(x)yΒΉ + sin(x)y = 2cosΒ³(x)sin(x) - 1; y(pi/4) =3/2, 0<x<pi/2find the value of b for which the given equation is exact, and then solve it using that value of b
(xy^(2)+bx^(2)y)dx+(x+y)x^(2)dy=0
Solve yβ=y-x2
, y(0)=1, by Picordβs method upto the third approximation . Hence find
the value of y(0,1), y(0,2)
solve the D.E dy/dx = (6x^5-2x+1)/(cosy+e^y)