What is the answer to solvefor a,y=ae^{2x}+be^{3x}+x^2 ?

The answer to solvefor a,y=ae^{2x}+be^{3x}+x^2 is a=(y-be^{3x}-x^2)/(e^{2x)}

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solvefor a,y=ae^{2x}+be^{3x}+x^2

solve for

a, y = a e^{2 x} + b e^{3 x} + x^{2}

a = \frac{y - b e ^{3 x} - x ^{2}}{e ^{2 x}}

Solution steps

y = a e^{2 x} + b e^{3 x} + x^{2}

Switch sides

a e^{2 x} + b e^{3 x} + x^{2} = y

Move

b e^{3 x}

to the right side

a e^{2 x} + x^{2} = y - b e^{3 x}

Move

x^{2}

to the right side

a e^{2 x} = y - b e^{3 x} - x^{2}

Divide both sides by

e^{2 x}

a = \frac{y - b e ^{3 x} - x ^{2}}{e ^{2 x}}

x, x - 3 y - 6 = 0

m, p = m + 28

m, E = \frac{1}{2} m v^{2}

f, f = \frac{1}{x ^{2} + y ^{2} - 4}

- \frac{5}{11} h + \frac{7}{9} = \frac{2}{9}

What is the answer to solvefor a,y=ae^{2x}+be^{3x}+x^2 ?
The answer to solvefor a,y=ae^{2x}+be^{3x}+x^2 is a=(y-be^{3x}-x^2)/(e^{2x)}