Question #da059

#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)#

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)#

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)#