What is (y'(t)+y)^2(L-y'(t)-y)^2 ?

The solution to (y'(t)+y)^2(L-y'(t)-y)^2 is (y+y'(t))^2(L-y-y'(t))^2

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simplify (y'(t)+y)^2(L-y'(t)-y)^2

simplify

(y^{'} (t) + y)^{2} (L - y^{'} (t) - y)^{2}

(y + y' (t))^{2} (L - y - y' (t))^{2}

Solution steps

(y^{'} (t) + y)^{2} (L - y^{'} (t) - y)^{2}

Apply exponent rule:

a^{m} b^{m} = (ab)^{m},

assuming

a \geq 0, b \geq 0

(y^{'} (t) + y)^{2} (L - y^{'} (t) - y)^{2} = ((y^{'} (t) + y) (L - y^{'} (t) - y))^{2}

= ((y^{^{'}} (t) + y) (L - y^{^{'}} (t) - y))^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= (y^{^{'}} (t) + y)^{2} (L - y^{^{'}} (t) - y)^{2}

x^{2} + y^{2} \cdot 16 - x^{2} - y^{2}

(3 x^{\frac{1}{3}}) (4 x^{\frac{1}{6}})

e^{- 3.36}

a^{2} + b^{2} c^{2} + d^{2}

6.3 \cdot 1 0^{- 8}

What is (y'(t)+y)^2(L-y'(t)-y)^2 ?
The solution to (y'(t)+y)^2(L-y'(t)-y)^2 is (y+y'(t))^2(L-y-y'(t))^2