What are the solutions to the equation solvefor s,t(st^2+2t+s)ds+s^2(1-t^2)dt=0 ?

The solutions to the equation solvefor s,t(st^2+2t+s)ds+s^2(1-t^2)dt=0 are s=0,s=-t;\quad 2dt\ne 0

Find the zeros of solvefor s,t(st^2+2t+s)ds+s^2(1-t^2)dt=0

The zeros of solvefor s,t(st^2+2t+s)ds+s^2(1-t^2)dt=0 are s=0,s=-t;\quad 2dt\ne 0

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solve for

s, t (s t^{2} + 2 t + s) d s + s^{2} (1 - t^{2}) d t = 0

s = 0, s = - t; 2 d t \neq = 0

Solution steps

t (s t^{2} + 2 t + s) d s + s^{2} (1 - t^{2}) d t = 0

Expand

t (s t^{2} + 2 t + s) d s + s^{2} (1 - t^{2}) d t : 2 d s t^{2} + 2 d s^{2} t

2 d s t^{2} + 2 d s^{2} t = 0

Write in the standard form

a x^{2} + b x + c = 0

2 d t s^{2} + 2 d t^{2} s = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- 2 d t ^{2} \pm ( 2 d t ^{2} ) ^{2} - 4 \cdot 2 d t \cdot 0}{2 \cdot 2 d t}

Simplify

(2 d t^{2})^{2} - 4 \cdot 2 d t \cdot 0 : 2 d t^{2}

s_{1, 2} = \frac{- 2 d t ^{2} \pm 2 d t ^{2}}{2 \cdot 2 d t}; 2 d t \neq = 0

Separate the solutions

s_{1} = \frac{- 2 d t ^{2} + 2 d t ^{2}}{2 \cdot 2 d t}, s_{2} = \frac{- 2 d t ^{2} - 2 d t ^{2}}{2 \cdot 2 d t}

s = \frac{- 2 d t ^{2} + 2 d t ^{2}}{2 \cdot 2 d t} : 0

s = \frac{- 2 d t ^{2} - 2 d t ^{2}}{2 \cdot 2 d t} : - t

The solutions to the quadratic equation are:

s = 0, s = - t; 2 d t \neq = 0

x, x^{2} + y^{2} + 2 x + 2 y + 34 = 0

(3 x - 2) (x + 3) + (x - 2) (x + 5) = - 2 (x + 8)

32 = - 16 t^{2} + 48 t

24 x^{2} + 36 x - 24 = 0

3 x^{2} + 5 x + 2 = 0

What are the solutions to the equation solvefor s,t(st^2+2t+s)ds+s^2(1-t^2)dt=0 ?
The solutions to the equation solvefor s,t(st^2+2t+s)ds+s^2(1-t^2)dt=0 are s=0,s=-t;\quad 2dt\ne 0
Find the zeros of solvefor s,t(st^2+2t+s)ds+s^2(1-t^2)dt=0
The zeros of solvefor s,t(st^2+2t+s)ds+s^2(1-t^2)dt=0 are s=0,s=-t;\quad 2dt\ne 0