Help solving the differential equation dy/dx+6y=7x?

Question

with the initial condition y(0)=2 obtain the exact solution of the differential equation

Answer 1

First, let's solve dy/dx + 6y =0, a first order linear differential equation.

Then dy/y = - 6 dx => ln(y) = -6x + c, c a constant. Therefore, y = ke^(-6x), wher k = e^c is a constant.

The solution of dy/dx+6y=7x is composed of 2 parts: the homogeneous sulution yh, which is the solution of the previous equation, and a particular solution yp to be determined. So,

y = yh + yp = ke^(-6x) + yp. Since the right hand side is 7x, a polynomial function of degree 1, yp is a polynomial of degree 1 too. So, yp = ax + b where a, and b are coefficientes to determine. Substituting yp n the equation gives

a + 6ax |+ 6b = 7x , so that 6a = 7 and a + 6b =0.

It follows a = 7/6 and b = -a/6 = 7/36 and yp = (7/6) x - 7/36.

And our solution is y = ke^(-6x) + (7/6) x - 7/36

Since y(0) = 2, it follows ke^(0) - 7/36 => k - 7/36 = 2 =>. k = 7/36 + 2 = 79/36
Our solution is then y = (79/36) e^(-6x) + (7/6) x - 7/36

Answer 2

Take the equation without second member
dy/dx+6y=0
dy/y=-6dx so lnIyI= -6x+C
y= Ke^-6x
Let´s find a particulat solution of the equation with 2nd side
y=mx+n

m +6mx+6n= 7x
6m=7 m= 7/6 and m+6n=0 so n= -7/36
The general solution is
y=Ke^-6x +7/6x-7/36
If we put x=0
2= K -7/36 so K = 79/36

y=79/36 e^-6x +7/6x-7/36

Answer 3

dy/dx = 7x-6y

let y = 7/6(x+t) then t = 6/7 y -x

dy/dx = 7/6 + 7/6 dt/dx

or 7/6 + 7/6 dt/dx = 7x - 7(x+t) = - 7t

multiply by 6
7+ 7 dt/dx = -42 t
devide by 7

1+ dt/dx = -6t
do dt/dx = 1+ 6t
or dt/(1+6t) = dx

integarting ln(1+6t) = 6x

or 1+6t = e^(6x)
or 1+6(6/7y-x) = ce^*(6x)

when you put x = 0 you get 1+36/7*2 = c or c 25/7

we get the equation
1+6(6/7y-x) = 25/7e^(6x)

Answer 4

Y'' +3/t y' = 2/t^2 enable y'= u u' + 3/t u = 2/t^2 -------------------(a million) Linear equation of first order. U' +p(t) u = q(t) p(t) = 3/t q(t) = 2/t^2 locate the integrating component e^? p(t) dt = e^? 3/t dt = e^ 3 ln(t) = e^ ln(t^3) = t^3 Multiply (a million) by skill of skill of t^3 u' t^3 + 3t^2 u = 2t --------------(2) (ut^3)' = 2t combine the two aspects ut^3 = 2t^2/2 + C u t^3 = t^2 + C u = a million/t+ C/t^3 substitute u with y' y' = a million/t + C/t^3 combine the two aspects y = ln(t) -C/2t^2 + C1