Question #da059

3 Answers
Jan 8, 2017

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)

Jan 8, 2017

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)

Jan 8, 2017

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)