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dy/dx = -2x + 3, y(3) = 3

A) y = x^{2} + 3x + 1

B) y = x^{2} + 2x - 2

C) y = -x^{2} + 3x + 3

D) y = x^{2} + 2x + 1

Asked by judziy•rep:213

University of Benin • 25 June 2020

University of Benin • 25 June 2020

**Comments**

Peter answered•rep:202

School not set • 26 June 2020

School not set • 26 June 2020

0

dy/dx = -2x + 3

Multiply both sides by dx,

dy = (-2x - 1) dx

Integrating both sides,

∫dy = ∫(-2x + 3) dx

y = -2x^{2}/2 + 3x + c

y = - x^{2} + 3x + c ------------ eqn(1)

Where c is the integration constant.

Since y(3) = 3,

3 = - (3)^{2} + 3(3) + c

3 = - 9 + 9 + c

3 = 0 + c

c = 3

Thus, eqn(1) becomes:

y = - x^{2} + 3x + 3

**Ans C**

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