Question

Question #da059

Answer 1

The general solution is `y=e^x/(a+1)+Ce^(-ax)`

Explanation:

`dy/dx+ay=e^x`

Multiply both sides by the integrating factor

`=e^(intadx)=e^(ax)`

`e^(ax)dy/dx+e^(ax)ay=e^(ax)*e^x`

`d/dx(ye^(ax))=e^(ax+x)`

Integrating both sides

`ye^(ax)=inte^(ax+x)dx=e^(x(a+1))/(a+1)+C`

Dividing both sides by `e^(ax)`

`y=e^(x(a+1))/(e^(ax)(a+1))+Ce^(-ax)`

`y=e^x/(a+1)+Ce^(-ax)`

Answer 2

`y = C_1e^(-ax)+e^x/(a+1)`

Explanation:

This is a non-homogeneous linear differential equation. The solution can be obtained as the sum of a particular solution `y_p` plus the homogeneous solution `y_h`.

The homogeneous solution obeys

`(dy_h)/(dx)+ay_h=0` This equation is separable giving

`(dy_h)/y_h=-adx` and the solution is

`logy_h= -ax +C->y_h=C_1e^(-ax)`

The particular solution is obtained with the "Constants Variation" method due to Lagrange.

Supposing `y_p=C_1(x)e^(-ax)` and substituting into

`(dy_p)/(dx) + ay_p = e^x` we obtain

`C_1'(x)e^(-ax)=e^x` then

`C_1(x)=e^((a+1)x)/(a+1)`

Finally the solution is

`y=y_h+y_p=C_1e^(-ax)+e^((a+1)x)/(a+1)e^(-ax)=C_1e^(-ax)+e^x/(a+1)`

Answer 3

`y(x) = e^x/(a+1)+Ce^(-ax)`

Explanation:

This is a linear differential equation, so first we have to find the integrating factor:

`u(x) =e^(int adx) = e^(ax)`

We have then that:

`d/(dx) (e^(ax) y(x)) = e^(ax) (dy)/(dx) +ae^(ax)y(x) = e^(ax)((dy)/(dx) +ay(x))`

Now we start from the original equation, multiply both sides by the integrating factor and use the identity here above:

`(dy)/(dx) +ay(x) = e^x`

`e^(ax)((dy)/(dx) +ay(x)) = e^(ax)e^x =e^((a+1)x)`

`d/(dx) (e^(ax) y(x))=e^((a+1)x)`

Now we can separate variables:

`e^(ax)y(x) = int e^((a+1)x)dx`

`e^(ax)y(x) = 1/(a+1)e^((a+1)x)+C`

`y(x) = e^(-ax)/(a+1)e^((a+1)x) +Ce^(-ax) = e^x/(a+1)+Ce^(-ax)`