0

dy/dx = 4x + 2, y(0) = 0

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

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

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

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

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 + 2

Multiply both sides by dx,

dy = (4x + 2) dx

Integrating both sides,

∫dy = ∫(4x + 2) dx

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

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

Where c is the integration constant.

Since y(0) = 0,

0 = 2(0)^{2} + 2(0) + c

0 = 0 + c

c = 0

Thus, eqn(1) becomes:

y = 2x^{2} + 2x

**Ans A**

Your Answer

**JMAP**

2 followers

141 questions

**math**

4 followers

843 questions

**differential-equation**

1 followers

8 questions