Processing math: 100%

Solution for dx/dy + px = q

Solution for dx/dy + px = q, where p and q are functions of y or constants

A differential equation is called linear if there are no multiplications among dependent variables and their derivatives. In other words, all coefficients are functions of independent variables. It can be written in two forms. The second is discussed in detail here.

dxdy+Px=Q

Such that, P and Q are either constants or functions of y variable only. This type of differential equation is called a first-order linear differential equation. To solve this type of equation, we first multiply both the sides of the equation by a function of y: g(y):

g(y)dxdy+P(g(y))x=Qg(y)

The function g(x) is chosen such that RHS becomes the derivative of y.g(x)

Therefore,
g(y)dxdy+P(g(y))x=g(y)dxdy+xgโ€˜(y)

P=gโ€˜(y)g(y)

Now we integrate both the sides with respect to x to obtain:
โˆซPdy=โˆซgโ€˜(y)g(y)dy

โˆซPdy=log(g(y))

Therefore:

g(y)=eโˆซPdy

eโˆซPdydxdy+PeโˆซPdyx=QeโˆซPdy

ddx(xeโˆซPdy)=QeโˆซPdy

By integrating both equations we obtain:

xeโˆซPdy=โˆซ(QeโˆซPdy)dy

y=eโˆ’โˆซPdyโ‹…โˆซQeโˆซPdydy+C

Steps used to solve first-order linear differential equation are as follows:

(i) Write the given differential equation in the following first-order linear DE form dxdy+Px=Q

where P, Q are constants or functions of y only.

(ii) Find the Integrating Factor (IF): eโˆซPdy

(iii) Write the solution of the given differential equation using IF in the following manner: x(IF)=โˆซ(Qร—IF)dy+C

Example: Find the general solution of the linear DE: ydxโˆ’(x+2y2)dy=0

This equation can be written in the form: dxdyโˆ’xy=2y

This is a linear differential equation of the type discussed above: dxdy+Px=Q

Here P = -1/y and Q = 2y which are both functions of y.

Therefore,
IF=eโˆซPdy=eโˆซโˆ’1ydy=eโˆ’logy=1y

Hence, the solution of the given differential equation can be written as:

x1y=โˆซ2y(1y)dy+c

Or

xy=โˆซ2dy+c

Integrating the equation gives:

xy=2y+c

Hence we can write:

x=2y2+cy

Since c is also a function of y. This is the general solution of the linear DE.

Example: Solve the given differential equation: (tanโˆ’1yโˆ’x)dy=(1+y2)dx

We can write this equation in a linear form, such that: dxdy+x1+y2=tanโˆ’1y1+y2

In this case, we identify that P = 11+y2 and Q = tanโˆ’1y1+y2 where P and Q are both functions of y

Hence we can write the equation using the integrating factor:

IF=eโˆซ11+y2dy=etanโˆ’1y

Now we can write the given differential equation is:

xetanโˆ’1y=โˆซtanโˆ’1y1+y2etanโˆ’1ydy+c

Now let S = โˆซtanโˆ’1y1+y2etanโˆ’1ydy

Substituting tanโˆ’1y=t, such that 11+y2dy=dt

Now, we get:

S = โˆซtanโˆ’1y1+y2etanโˆ’1ydy=โˆซtetdt

=tetโˆ’โˆซ1etdt=tetโˆ’et=et(tโˆ’1)

S=etanโˆ’1y(tanโˆ’1yโˆ’1)

Now we substitute the value of S in the differential equation:

xetanโˆ’1y=etanโˆ’1y(tanโˆ’1yโˆ’1)+c

Hence the general solution for the differential equation is:

x=(tanโˆ’1y)+cetanโˆ’1y

 

Content Protection by DMCA.com
Please Share