0

dy/dx = -2x - 1, y(-1) = -1

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

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

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

D) y = -x^{2} - x - 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 • 25 June 2020

School not set • 25 June 2020

0

dy/dx = -2x - 1

Multiply both sides by dx,

dy = (-2x - 1) dx

Integrating both sides,

∫dy = ∫(-2x - 1) dx

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

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

Where c is the integration constant.

Since y(-1) = -1,

-1 = - (-1)^{2} - (-1) + c

-1 = - 1 + 1 + c

-1 = 0 + c

c = -1

Thus, eqn(1) becomes:

y = - x^{2} - x - 1

**Ans D**

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