To solve the given partial differential equation (PDE), we can use the method of characteristics.
First, we need to find the characteristic equations associated with the given PDE. The characteristic equations are given by:
dx / dt = y + u
dy / dt = y
du / dt = x - y
Next, we solve these equations simultaneously to determine the characteristics.
From the second equation, we can see that dy / dt = y, which has the solution y = Ce^t, where C is a constant.
Differentiating the first equation with respect to t, we have d^2x / dt^2 = du / dt. Substituting the third equation into this, we get d^2x / dt^2 = x - y.
Since y = Ce^t, we have d^2x / dt^2 = x - Ce^t. This is a linear second-order ordinary differential equation with constant coefficients. We can solve this equation using the characteristic equation method:
Assume the solution has the form x = Ae^t + Be^(-t), where A and B are constants.
Differentiating x with respect to t, we have dx / dt = Ae^t - Be^(-t).
Substituting this into the equation d^2x / dt^2 = x - Ce^t, we get (Ae^t - Be^(-t)) - Ce^t = Ae^t + Be^(-t) - Ce^t.
Simplifying this equation, we find A - C = A + B - C.
Thus, A = B, and the solution for x is x = Ae^t + Ae^(-t) = A(e^t + e^(-t)), where A is a constant.
Now, we can solve for u using the first equation:
dx / dt = y + u
(Ae^t + Ae^(-t))' = Ce^t + u
Differentiating both sides of the equation, we get:
(Ae^t - Ae^(-t))' = Ce^t + u
Ae^t + Ae^(-t) = Ce^t + u
We can substitute y = Ce^t into this equation, giving us:
(Ae^t + Ae^(-t)) = y + u
Since (Ae^t + Ae^(-t)) = x, we have:
x = y + u
Therefore, the general solution to the given PDE is u(x, y) = x - y.
To find the particular solution, we use the initial condition u(x, 1) = 1 + x.
Substituting y = 1 into the general solution, we get:
u(x, 1) = x - 1
Since u(x, 1) = 1 + x is equal to x - 1, the particular solution to the given PDE is u(x, y) = x - 1.
Solve below PDE
(y+u)*u_x + y*u_y = x - y , u(x, 1) = 1+x
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