Chapter Two Linear First-Order Differential Equations

Chapter Two
Linear First-Order Differential Equations
The standard form for the linear first order diff. eq. is ;
y^ +p(x)y=q(x)
Or
dy/dx+p(x)y=q(x)
Where p(x) ,q(x) are continuous functions ,is linear in y ,because y and its derivative dy/dx occur only to the first power are not multiplied together, nor do they appear as the argument of a function (such as sin y , e^y or ?(dy/dx)

For example all these are non linear


to solve the linear eq. Find the integrating factor I(x) as ;
I(x)=e^??p(x)dx
Which depends only on x and is independent of y ,and multiply both sides by I(x) the result is;
I(x)y^ +I(x)p(x)y=I(x)q(x)
Is exact , Which can solved by usual method, or rewrite as ;
(d(yI(x)))/dx=I(x)q(x)
Example: put the following eq. in standard form ;

Solution ; Rewrite eq. as


Example: Find an integrating factor for eq. ; y^ -3y=6 and then find the solution .
Solution; This diff. Eq. is linear with p(x)=-3 , q(x)=6
Here I(x)=e^??p(x)dx=e^???-3dx?=e^(-3x)
To solve this eq. multiply by integrating factor as;
?e^(-3x) y?^ -3e^(-3x) y=6e^(-3x)
Or d/dx(ye^(-3x))=6e^(-3x)
Integrating both sides with respect to x
ye^(-3x)=???6e^(-3x) dx?
=-2e^(-3x)+c
y=ce^3x-2

Note: Some times the variables may be change with respect to independent x and depends on y ,to be
written as;
dx/dy+g(y)x=h(y)
and the integrating factor is e^??g(y)dy .