What is the general solution for solvefor y,xy+2y=sin(x) ?

The general solution for solvefor y,xy+2y=sin(x) is y=(sin(x))/(x+2);\quad x\ne-2

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solvefor y,xy+2y=sin(x)

solve for

y, x y + 2 y = sin (x)

y = \frac{sin ( x )}{x + 2}; x \neq = - 2

Solution steps

x y + 2 y = sin (x)

Factor

x y + 2 y : y (x + 2)

y (x + 2) = sin (x)

Divide both sides by

x + 2; x \neq = - 2

y = \frac{sin ( x )}{x + 2}; x \neq = - 2

cos (1 5^{\circ}) = \frac{a}{55}

tan^{0} (2 x + 15) = - 1

x = \frac{20}{sin ( 4 5 ^{\circ} )}

sin^{4} (u) - cos^{4} (u) = 2 sin^{2} (u) - 1

tan^{4} (x) - sec^{4} (x) = - 2 sec^{2} (x) + 1

What is the general solution for solvefor y,xy+2y=sin(x) ?
The general solution for solvefor y,xy+2y=sin(x) is y=(sin(x))/(x+2);\quad x\ne-2