What is t(x)x^{''}+(4t(x)-2)x^'+4x ?

The solution to t(x)x^{''}+(4t(x)-2)x^'+4x is tx^{1+'^2}+x^'(4tx-2)+4x

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simplify t(x)x^{''}+(4t(x)-2)x^'+4x

simplify

t (x) x^{^{''}} + (4 t (x) - 2) x^{^{'}} + 4 x

t x^{1 +'^{2}} + x^{'} (4 t x - 2) + 4 x

Solution steps

t (x) x^{^{''}} + (4 t (x) - 2) x^{^{'}} + 4 x

Remove parentheses:

(a) = a

= t x x^{^{^{''}}} + (4 t x - 2) x^{^{^{'}}} + 4 x

\frac{d}{d y} ((x) (2 x + 1))

\frac{sec ^{2} ( x ) ( x ^{2} + 1 ) ^{6}}{( e ^{x} - cos ^{2} ( x ) ) ^{4}}

1 + t^{2} \cdot (1 + t^{2})

(2 x - y) (2 x + y)

d x (- \frac{7}{3 x ^{3}} + x + \frac{5}{x})

What is t(x)x^{''}+(4t(x)-2)x^'+4x ?
The solution to t(x)x^{''}+(4t(x)-2)x^'+4x is tx^{1+'^2}+x^'(4tx-2)+4x