First Order Differential Equation

[Woolyabyss]

Homework Statement

Find the general solution to the following differential equation.

dy/dx = 2x( (y^2) + 1)

Homework Equations

The Attempt at a Solution

I got all x terms on one side and all y terms on the otherside

2x dx = 1/( (y^2) + 1`)dy

integrate

x^2 + c = tan(y)

y = tan-inverse(x^2 + c)

Can somebody tell me if this is right ?

my book says the solution is y = tan(x^2 + c)

 

[phyzguy]

It's not right. What is
$$\int \frac{dy}{1+y^2}$$?

 

[Woolyabyss]

tan-inverse(1/y)

sorry I see where I went wrong now

it should have been x^2 + c = tan-inverse(y)

 

[Tanya Sharma]

tan-inverse(1/y)

sorry I see where I went wrong now

it should have been x^2 + c = tan-inverse(y)

∫dy/(1+y²) = tan^-1y

So,tan^-1y = x² + c

or , y = tan(x² + c)

 

[Woolyabyss]

∫dy/(1+y²) = tan^-1y

So,tan^-1y = x² + c

or , y = tan(x² + c)

Could you help me with another one?

Given the differential equation

x(dy/dx) = √(4 - y^2) , find the general solution given that y = 0 when x = 1

I got all the x terms to one side and all of the y terms to the opposite side.

1/√(4 - y^2) dy = 1/x dx

integrating gives

sin-inverse(y/2) = lnx

using the limits and we get

sin-inverse(y/2) = lnx

y/2 = sin(lnx)

y = 2sin(lnx)

my book says the solution is y = sin(lnx)

 

[Tanya Sharma]

Could you help me with another one?

Given the differential equation

x(dy/dx) = √(4 - y^2) , find the general solution given that y = 0 when x = 1

I got all the x terms to one side and all of the y terms to the opposite side.

1/√(4 - y^2) dy = 1/x dx

integrating gives

sin-inverse(y/2) = lnx

using the limits and we get

sin-inverse(y/2) = lnx

y/2 = sin(lnx)

y = 2sin(lnx)

my book says the solution is y = sin(lnx)

y=2sin(lnx) looks to be the correct answer.You can verify that by putting the solution in the original equation and checking whether it satisfies the eq.

Note : Even though you came up with the correct answer ,please do not ignore the constant of integration C while integrating.