What is the answer to solvefor A,Y=Ae^{2x}+Bxe^{2x} ?

The answer to solvefor A,Y=Ae^{2x}+Bxe^{2x} is A=(Y-Be^{2x}x)/(e^{2x)}

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solvefor A,Y=Ae^{2x}+Bxe^{2x}

solve for

A, Y = A e^{2 x} + B x e^{2 x}

A = \frac{Y - B e ^{2 x} x}{e ^{2 x}}

Solution steps

Y = A e^{2 x} + B x e^{2 x}

Switch sides

A e^{2 x} + B x e^{2 x} = Y

Move

B x e^{2 x}

to the right side

A e^{2 x} = Y - B x e^{2 x}

Divide both sides by

e^{2 x}

A = \frac{Y - B e ^{2 x} x}{e ^{2 x}}

7 - 3 n = - 2 n

4 w + 20 = 0

- 8 y + 1 = - 9 y

x, 4 x - y = 0

x, 4 x - y = 7

What is the answer to solvefor A,Y=Ae^{2x}+Bxe^{2x} ?
The answer to solvefor A,Y=Ae^{2x}+Bxe^{2x} is A=(Y-Be^{2x}x)/(e^{2x)}