What is a solution to the differential equation (1+y)dy/dx-4x=0(1+y)dydx−4x=0?

2 Answers
Narad T.
Jul 29, 2017

The solution is y=+-sqrt(2x^2+1+C_1)-1y=±√2x2+1+C1−1

Explanation:

(1+y)dy/dx-4x=0(1+y)dydx−4x=0

This is a first order ordinary differential equation of the form

N(y)dy=M(x)dxN(y)dy=M(x)dx

Therefore,

(1+y)dy/dx=4x(1+y)dydx=4x

(1+y)dy=4xdx(1+y)dy=4xdx

Integrating both sides

int(1+y)dy=int4xdx∫(1+y)dy=∫4xdx

y+y^2/2=4x^2/2+Cy+y22=4x22+C

y^2/2+y=2x^2+Cy22+y=2x2+C

y^2+2y=2x^2+C_1y2+2y=2x2+C1

y^2+2y+1=2x^2+1+C_1y2+2y+1=2x2+1+C1

(y+1)^2=2x^2+1+C_1(y+1)2=2x2+1+C1

(y+1)=+-sqrt(2x^2+1+C_1)(y+1)=±√2x2+1+C1

y=+-sqrt(2x^2+1+C_1)-1y=±√2x2+1+C1−1

Cesareo R.
Jul 29, 2017

See below.

Explanation:

1/2(d/(dx))(1+y)^2-2(d/(dx))x^2=(d/dx)(1/2(1+y)^2-2x^2)=012(ddx)(1+y)2−2(ddx)x2=(ddx)(12(1+y)2−2x2)=0

then

1/2(1+y)^2-2x^2=C12(1+y)2−2x2=C or

(1+y)^2=4x^2+C_1(1+y)2=4x2+C1 or

1+y=pmsqrt(4x^2+C_1)1+y=±√4x2+C1 or

y = -1pmsqrt(4x^2+C_1)y=−1±√4x2+C1

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