How to you find the general solution of #dy/dx=x/(1+x^2)#?

Answer 1

Multiply both sides by #dx# and then integrate.

Given: #dy/dx=x/(1+x^2)#
Multiply both sides by #dx#
#dy=x/(1+x^2)dx#

Integrate:

#intdy=intx/(1+x^2)dx#
#intdy=1/2int(2x)/(1+x^2)dx#
#intdy = 1/2(du)/u#
#y = 1/2ln(u)+ C#
#y = 1/2ln(x^2+1)+C#
#y = ln(sqrt(x^2+1))+C#
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Answer 2

To find the general solution of the differential equation ( \frac{dy}{dx} = \frac{x}{1+x^2} ), you can follow these steps:

  1. Separate the variables by moving all terms involving ( y ) to one side and all terms involving ( x ) to the other side.
  2. Integrate both sides with respect to their respective variables.
  3. Solve for ( y ) to find the general solution.

Here's how to do it:

[ \frac{dy}{dx} = \frac{x}{1+x^2} ]

Separate variables:

[ \frac{dy}{1} = \frac{x}{1+x^2} , dx ]

Integrate both sides:

[ \int , dy = \int \frac{x}{1+x^2} , dx ]

[ y = \int \frac{x}{1+x^2} , dx ]

To integrate ( \frac{x}{1+x^2} ), perform a substitution. Let ( u = 1 + x^2 ), then ( du = 2x , dx ), so ( \frac{1}{2} du = x , dx ).

Substitute ( u ) and ( \frac{1}{2} du ) back into the integral:

[ y = \int \frac{1}{u} , \frac{1}{2} du ]

[ y = \frac{1}{2} \int \frac{1}{u} , du ]

[ y = \frac{1}{2} \ln|u| + C ]

Substitute ( u = 1 + x^2 ) back into the equation:

[ y = \frac{1}{2} \ln|1 + x^2| + C ]

So, the general solution to the differential equation is:

[ y = \frac{1}{2} \ln|1 + x^2| + C ]

where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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