What is the answer to solvefor x,(d^2-2d+4)y=e^cosx ?

The answer to solvefor x,(d^2-2d+4)y=e^cosx is x=((d^2-2d+4)y)/(e^cos);\quad e^cos\ne 0

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solvefor x,(d^2-2d+4)y=e^cosx

solve for

x, (d^{2} - 2 d + 4) y = e^{c} os x

x = \frac{( d ^{2} - 2 d + 4 ) y}{e ^{c} os}; e^{c} os \neq = 0

Solution steps

(d^{2} - 2 d + 4) y = e^{c} os x

Switch sides

e^{c} os x = (d^{2} - 2 d + 4) y

Divide both sides by

e^{c} os; e^{c} os \neq = 0

x = \frac{( d ^{2} - 2 d + 4 ) y}{e ^{c} os}; e^{c} os \neq = 0

x, 2 y = 2 x + 1.2

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x, 150 x + 250 y = 55000

3 x + 2 = \frac{5 x}{8 + 4}

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What is the answer to solvefor x,(d^2-2d+4)y=e^cosx ?
The answer to solvefor x,(d^2-2d+4)y=e^cosx is x=((d^2-2d+4)y)/(e^cos);\quad e^cos\ne 0