0

dy/dx = 4x - 3, y(1) = 0

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

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

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

D) y = 2x^{2} - 3x + 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 = 4x - 3

Multiply both sides by dx,

dy = (4x - 3) dx

Integrating both sides,

∫dy = ∫(4x - 3) dx

y = 4x^{2}/2 - 3x + c

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

Where c is the integration constant.

Since y(1) = 0,

0 = 2(1)^{2} - 3(1) + c

0 = 2 - 3 + c

0 = -1 + c

Add 1 to both sides,

c = 1

Thus, eqn(1) becomes:

y = 2x^{2} - 3x + 1

**Ans D**

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