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

(1+y)dydx−4x=0This is a first order ordinary differential equation of the formN(y)dy=M(x)dxTherefore,(1+y)dydx=4x(1+y)dy=4xdxIntegrating both sides∫(1+y)dy=∫4xdxy+y22=4x22+Cy22+y=2x2+Cy2+2y=2x2+C1y2+2y+1=2x2+1+C1(y+1)2=2x2+1+C1(y+1)=±2x2+1+C1y=±2x2+1+C1−1

12(ddx)(1+y)2−2(ddx)x2=(ddx)(12(1+y)2−2x2)=0then12(1+y)2−2x2=C or(1+y)2=4x2+C1 or1+y=±4x2+C1 ory=−1±4x2+C1

(1+y)dydx−4x=0This is a first order ordinary differential equation of the formN(y)dy=M(x)dxTherefore,(1+y)dydx=4x(1+y)dy=4xdxIntegrating both sides∫(1+y)dy=∫4xdxy+y22=4x22+Cy22+y=2x2+Cy2+2y=2x2+C1y2+2y+1=2x2+1+C1(y+1)2=2x2+1+C1(y+1)=±2x2+1+C1y=±2x2+1+C1−1

12(ddx)(1+y)2−2(ddx)x2=(ddx)(12(1+y)2−2x2)=0then12(1+y)2−2x2=C or(1+y)2=4x2+C1 or1+y=±4x2+C1 ory=−1±4x2+C1

Solved: "What is a solution to the differential equation..."

High School
Calculus and Analysis
Calculus 2
Applications of integrals

Emmy Decker

Answered question

2022-04-16

What is a solution to the differential equation

(1 + y) \frac{d y}{d x} - 4 x = 0

Answer & Explanation

abangan85s0

Beginner2022-04-17Added 16 answers

(1 + y) \frac{d y}{d x} - 4 x = 0

This is a first order ordinary differential equation of the form
N(y)dy=M(x)dx
Therefore,

(1 + y) \frac{d y}{d x} = 4 x

(1+y)dy=4xdx
Integrating both sides

\int (1 + y) d y = \int 4 x d x

y + \frac{y^{2}}{2} = 4 \frac{x^{2}}{2} + C

\frac{y^{2}}{2} + y = 2 x^{2} + C

y^{2} + 2 y = 2 x^{2} + C_{1}

y^{2} + 2 y + 1 = 2 x^{2} + 1 + C_{1}

{(y + 1)}^{2} = 2 x^{2} + 1 + C_{1}

(y + 1) = \pm \sqrt{2 x^{2} + 1 + C_{1}}

y = \pm \sqrt{2 x^{2} + 1 + C_{1}} - 1

ze2m1ingkdvu

Beginner2022-04-18Added 16 answers

\frac{1}{2} (\frac{d}{d x}) {(1 + y)}^{2} - 2 (\frac{d}{d x}) x^{2} = (\frac{d}{d x}) (\frac{1}{2} {(1 + y)}^{2} - 2 x^{2}) = 0

then

\frac{1}{2} {(1 + y)}^{2} - 2 x^{2} = C

{(1 + y)}^{2} = 4 x^{2} + C_{1}

1 + y = \pm \sqrt{4 x^{2} + C_{1}}

y = - 1 \pm \sqrt{4 x^{2} + C_{1}}

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