Question

Solve the differential equation\[yu_x+u_y=u\]\[u(x,0)=f(x)\]
I got:\[u(x,y)=f(x-\frac{y^2}{2})e^{\frac{y}{2}}\]Anyone care to check this is right?

Answer 1

how u solve it?

Answer 2

I believe the problem becomes easier if you divide the original equation by y, so we get:\[u_x+\frac{1}{y}u_y=\frac{1}{y}u\]

You assume that there is a function v(s,t) such that:\[v(s,t)=u(x(s,t),y(s,t))\]
Then notice that:\[v_t=u_xx_t+u_yy_t\]

Choose x_t and y_t so that they equal the coefficients in the original problem so;\[x_t=1\]\[y_t=\frac{1}{y}\]\[v_t=\frac{1}{y}v\]

Also since\[u(x,0)=f(x)\]We must have:\[x(s,0)=s\]\[y(s,0)=0\]\[v(s,0)=f(s)\]

We now have 3 ordinary differential equations and 3 conditions, solving them:
\[x_t=1\]\[x(s,t)=t+h(s)\]\[x(s,0)=s=h(s)\]\[x(s,t)=t+s\]
\[y_t=\frac{1}{y}\]\[y(s,t)=\sqrt{2t+2h(s)}\]\[y(s,0)=0=\sqrt{2h(s)}\]\[y(s,t)=\sqrt{2t}\]
\[v_t=\frac{1}{y}v\]\[v(s,t)=e^{\frac{t}{y}}h(s)\]\[v(s,0)=f(s)=h(s)\]\[v(s,t)=f(s)e^{\frac{t}{y}}\]

Plugging these 3 results back into\[v(s,t)=u(x(s,t),y(s,t))\]Gives:\[f(s)e^{\frac{t}{y}}=u(t+s,\sqrt{2t})\]

Using the three results from earlier, lets solve them for s and t. So that:\[t=\frac{y^2}{2}\]\[s=x-\frac{y^2}{2}\]

Now we can resubstitute s and t into the equation to put it back in terms of x and y.\[f(x-\frac{y^2}{2})e^{\frac{y}{2}}=u(x,y)\]

I hope did that right, I was slightly confused in certain parts. This method is called the method of characterstics, fyi.

Answer 3

*bookmark
i must learn this method

Answer 4

Yea, unfortunately I couldn't much good info about this method on the internet...

Answer 5

@eliassaab