# Differential Equations

• Jan 7th 2009, 02:39 PM
KayPee
Differential Equations
1.Obtain the solution to the differential equation

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

2. The first order differential equation

dy/dx-xy=x

has the solution y= ?

These are both separable.
1) $\displaystyle \frac{dy}{dx} = \frac{1 + y^2}{1 + x^2} \rightarrow \frac{dy}{1 + y^2} = \frac{dx}{1 + x^2}$

2) $\displaystyle \frac{dy}{dx} - xy = x \rightarrow \frac{dy}{dx} = x(1+y) \rightarrow \frac{dy}{1 + y} = x dx$

Someone kindly finish up for me..I kind of confused at this stage.
• Jan 7th 2009, 02:43 PM
Mush
Quote:

Originally Posted by KayPee
1.Obtain the solution to the differential equation

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

2. The first order differential equation

dy/dx-xy=x

has the solution y= ?

These are both separable.
1) $\displaystyle \frac{dy}{dx} = \frac{1 + y^2}{1 + x^2} \rightarrow \frac{dy}{1 + y^2} = \frac{dx}{1 + x^2}$

2) $\displaystyle \frac{dy}{dx} - xy = x \rightarrow \frac{dy}{dx} = x(1+y) \rightarrow \frac{dy}{1 + y} = x dx$

Someone kindly finish up for me..I kind of confused at this stage.

1) $\displaystyle \frac{dy}{1+y^2} = \frac{dx}{1+x^2}$

$\displaystyle \int \frac{dx}{x^2+a^2} = \frac{1}{a} \arctan{\frac{x}{a}} +C$

2) $\displaystyle \int \frac{dy}{1+y} = \ln{|1+y|}+C$
• Jan 10th 2009, 02:36 AM
KayPee
Quote:

Originally Posted by Mush
1) $\displaystyle \frac{dy}{1+y^2} = \frac{dx}{1+x^2}$

$\displaystyle \int \frac{dx}{x^2+a^2} = \frac{1}{a} \arctan{\frac{x}{a}} +C$

2) $\displaystyle \int \frac{dy}{1+y} = \ln{|1+y|}+C$

I would appreciate it if I could have the final answer.

I'm not getting it besides.I'm not used to the Latex
• Jan 10th 2009, 04:36 AM
Prove It
Quote:

Originally Posted by KayPee
1.Obtain the solution to the differential equation

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

2. The first order differential equation

dy/dx-xy=x

has the solution y= ?

These are both separable.
1) $\displaystyle \frac{dy}{dx} = \frac{1 + y^2}{1 + x^2} \rightarrow \frac{dy}{1 + y^2} = \frac{dx}{1 + x^2}$

2) $\displaystyle \frac{dy}{dx} - xy = x \rightarrow \frac{dy}{dx} = x(1+y) \rightarrow \frac{dy}{1 + y} = x dx$

Someone kindly finish up for me..I kind of confused at this stage.

You've been given the solution already, but I'll go step by step...

1. $\displaystyle \frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$

$\displaystyle \frac{1}{1 + y^2}\,\frac{dy}{dx} = \frac{1}{1 + x^2}$

$\displaystyle \int{\frac{1}{1 + y^2}\,\frac{dy}{dx}\,dx} = \int{\frac{1}{1 + x^2}\,dx}$

$\displaystyle \int{\frac{1}{1 + y^2}\,dy}= \int{\frac{1}{1 + x^2}\,dx}$

$\displaystyle \arctan{y} + C_1 = \arctan{x} + C_2$

$\displaystyle \arctan{y} = \arctan{x} + C$.

2. $\displaystyle \frac{dy}{dx} - xy = x$

$\displaystyle \frac{dy}{dx} = xy + x$

$\displaystyle \frac{dy}{dx} = x(y + 1)$

$\displaystyle \frac{1}{y + 1}\,\frac{dy}{dx} = x$

$\displaystyle \int{\frac{1}{y + 1}\,\frac{dy}{dx}\,dx} = \int{x\,dx}$

$\displaystyle \int{\frac{1}{y + 1}\,dy} = \int{x\,dx}$

$\displaystyle \ln{|y + 1|} + C_1 = \frac{1}{2}x^2 + C_2$

$\displaystyle \ln{|y + 1|} = \frac{1}{2}x^2 + C$

$\displaystyle |y + 1| = e^{\frac{1}{2}x^2 + C}$

$\displaystyle |y + 1| = e^C e^{\frac{1}{2}x^2}$

$\displaystyle y + 1 = \pm e^C e^{\frac{1}{2}x^2}$

$\displaystyle y + 1 = Ae^{\frac{1}{2}x^2}$

$\displaystyle y = Ae^{\frac{1}{2}x^2} - 1$.
• Jan 10th 2009, 04:41 AM
mr fantastic
Quote:

Originally Posted by Prove It
You've been given the solution already, but I'll go step by step...

1. $\displaystyle \frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$

$\displaystyle \frac{1}{1 + y^2}\,\frac{dy}{dx} = \frac{1}{1 + x^2}$

$\displaystyle \int{\frac{1}{1 + y^2}\,\frac{dy}{dx}\,dx} = \int{\frac{1}{1 + x^2}\,dx}$

$\displaystyle \int{\frac{1}{1 + y^2}\,dy}= \int{\frac{1}{1 + x^2}\,dx}$

$\displaystyle \arctan{y} + C_1 = \arctan{x} + C_2$

$\displaystyle \arctan{y} = \arctan{x} + C$.

2. $\displaystyle \frac{dy}{dx} - xy = x$

$\displaystyle \frac{dy}{dx} = xy + x$

$\displaystyle \frac{dy}{dx} = x(y + 1)$

$\displaystyle \frac{1}{y + 1}\,\frac{dy}{dx} = x$

$\displaystyle \int{\frac{1}{y + 1}\,\frac{dy}{dx}\,dx} = \int{x\,dx}$

$\displaystyle \int{\frac{1}{y + 1}\,dy} = \int{x\,dx}$

$\displaystyle \ln{|y + 1|} + C_1 = \frac{1}{2}x^2 + C_2$

$\displaystyle \ln{|y + 1|} = \frac{1}{2}x^2 + C$

$\displaystyle |y + 1| = e^{\frac{1}{2}x^2 + C}$

$\displaystyle |y + 1| = e^C e^{\frac{1}{2}x^2}$

$\displaystyle y + 1 = \pm e^C e^{\frac{1}{2}x^2}$

$\displaystyle y + 1 = Ae^{\frac{1}{2}x^2}$

$\displaystyle y = Ae^{\frac{1}{2}x^2} - 1$.

You missed a step (lol! Nice job with the latex, by the way).

To the OP:

1. Now, the least you can do in return is to confirm these solutions by substitution.

2. The solution given to you for the first question can be expressed in a simpler form .....
• Jan 10th 2009, 04:56 AM
Prove It
Quote:

Originally Posted by mr fantastic
You missed a step (lol! Nice job with the latex, by the way).

To the OP:

1. Now, the least you can do in return is to confirm these solutions by substitution.

2. The solution given to you for the first question can be expressed in a simpler form .....

I can't be bothered trying to get the first one to read $\displaystyle y = \dots$ because otherwise I'll be trying to find the most elegant way for it to read hahahaha.