Question

Solve the initial value problem (IVP): (y + x 2 y)y' = y^2 + 1, y(0) = 1

Answer #1

We solve the given differential equation by using seperation of variables method.

1.) 25pt) Solve the IVP: (initial value problem)
y’ = (3x2 + 4x + 2)/(2(y-1)), y(0) = -1

Consider the following Initial Value Problem (IVP)
y' = 2xy, y(0) = 1.
Does the IVP exists unique solution? Why? If it does, ﬁnd the
solution by Picard iteration with y0(x) = 1.

Solve the Initial Value Problem:
?x′ = 2y−x
y′ = 5x−y
Initial Conditions:
x(0)=2
y(0)=1

Solve the given initial-value problem.
y'' + x(y')2 = 0, y(1) = 5, y'(1) = 2 y =

y^4 - y = 0 IVP (Initial Value Problem)
y(0) = 0
y'(0) = 1
y''(0) = 0
y'''(0) = 0
The answer is y(t) = [sinh(t) + sin(t)] / 2 but I cannot figure
out how to reach this.

Solve the initial value problem
x′=−3x−y,
y′= 13x+y,
x(0) = 0,
y(0) = 1.

Solve the IVP: , y(0)=3.
2) Solve the DE: . y' = xy^2/ (x^2 +1)

solve the IVP
y'' - 4y' - 5y = 6e-x, y(0)= 1, y'(0) =
-2

solve the following initial value problem x' = 2xy
y' = y -2t +1 x(0) = x0 y(0) =
y0

Solve the 1st order initial value problem:
1+(x/y+cosy)dy/dx=0, y(pi/2)=0

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