The differential equation y '' - 3 / 2Y ^ 2 = 0 satisfies the initial y (0) = 1, y '(0) = 1

The differential equation y '' - 3 / 2Y ^ 2 = 0 satisfies the initial y (0) = 1, y '(0) = 1

Let y '= P (y), then y' '= DP / DX = DP / dy * P, the original equation is changed to 2pdp = 3Y ^ 2dy, ∧ p ^ 2 = y ^ 3 + C, ∧ y' = soil √ (y ^ 3 + C), substitute the initial condition into C = 0, take the positive sign, dy / √ (y ^ 3) = DX, - 2 / √ y = x + C1, substitute the initial condition into C1 = - 2, ∧ 2 / √ y = X-2, ∧ y = 2 / (2-x), ∧ y = [2 / (2-x)} ^ 2, (x

Solving differential equation 2Y '+ y = x / Y

Find the general solution of the differential equation y ″ - 2Y ′ + y = x + 1

The general solution of 1. Y ″ - 2Y ′ + y = 0
The characteristic equation is R & # 178; - 2R + 1 = 0
(r-1)²=0
r1=r2=1
Y=(c1+c2x)e^x
2. A special nonhomogeneous solution y*
Let y * = ax + B
y*'=a
y*''=0
-2a+ax+b=x+1
a=1,-2a+b=1
b=3
therefore
y*=x+3
therefore
General solution y = y + y * = (C1 + c2x) e ^ x + X + 3