Question

Solve the Differential Equation `dy/dx +3y = 0` with `x=0` when `y=1`?

Answer 1

`y = e^(-x)`

Explanation:

We have:

`dy/dx +3y = 0` with `x=0` when `y=1` ..... [A]

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

`dy/dx + P(x)y=Q(x)`

The given equation is already in standard form, So

Then the integrating factor is given by;

`I = e^(int P(x) dx)`
`\ \ = exp(int \ 3 \ dx)`
`\ \ = exp( 3x )`
`\ \ = e^(3x)`

And if we multiply the DE [A] by this Integrating Factor, `I`, we will have a perfect product differential;

`e^(3x)dy/dx +3e^(3x)y = 2e^(-x)e^(3x)`
`:. d/dx ( e^(3x)y ) = 2e^(2x)`

Which we can directly integrate to get:

`e^(3x)y = int \ 2e^(2x) \ dx`

`:. e^(3x)y = e^(2x) + C`

Using the initial condition `x=0` when `y=1`, we have:

`:. e^(0) = e^(0) + C => C = 0`

Thus,

`e^(3x)y = e^(2x)`

`:. y = e^(2x)e^(-3x)`
`\ \ \ \ \ \ \ = e^(-x)`

And the given answer is incorrect.

Answer 2

The solution is:

`y(x) = e^(-x)`

Explanation:

Solve the homogeneous differential equation:

`dy/dx +3y = 0`

The characteristic equation is:

`lambda + 3 = 0`

so the general solution of the homogeneous equation is:

`y_0(x) = Ce^(-3x)`

Search now a particular solution using Lagrange methods of variable coefficients in the form:

`y_1(x) = a(x) bary_0(x)`

where `bary_0(x)` is a non trivial solution of the homogeneous equation, and we choose the one for `C=1` so that:

`y_1(x) = a(x) e^(-3x)`

Substituting into the equation:

`(dy_1)/dx +3y_1 =2e^(-x)`

`d/dx( a(x) e^(-3x) ) +3 a(x) e^(-3x) = 2e^(-x)`

using the product rule:

`(da)/dx e^(-3x) -3 a(x) e^(-3x) + 3 a(x) e^(-3x) = 2e^(-x)`

`(da)/dx e^(-3x) = 2e^(-x)`

`(da)/dx = 2e^(2x)`

`a(x) = int 2e^(2x)dx = e^(2x) + c`

We can take the solution for `c=0` and have:

`y_1(x) = a(x)e^(-3x) = e^(2x)e^(-3x) = e^(-x)`

The general solution of the equation is then:

`y(x) = Ce^(-3x)+ e^-x`

Imposing `y(0) = 1` we get `C=0` and the required solution is:

`bary(x) = e^-x`

In fact:

`d/dx (e^(-x))+3e^(-x) = -e^(-x) +3e^(-x) = 2e^(-x)`

Answer 3

`y=e^(-3x)`

Explanation:

I try to find solution of `dy/dx+3y=0` differential equation with condition `y(0)=1`

After taking Laplace transformation both sides,

`sY(s)-y(0)+3Y(s)=0`

`(s+3)Y(s)-1=0`

`Y(s)=1/(s+3)`

After taking inverse Laplace transform, I found

`y=e^(-3x)`