Question

Show that u(x, y)=xf(2x+y) is a general solution of x∂u/∂x-2x ∂u/∂y=u

Answer 1

Answer on Question #68375 – Math – Differential Equations

Question

Show that $u(x, y) = xf(2x + y)$ is a general solution of $x \partial u / \partial x - 2x \partial u / \partial y = u$

Solution

Substitute

$$u(x, y) = xf(2x + y)$$

into the original equation $x \frac{\partial u}{\partial x} - 2x \frac{\partial u}{\partial y} = u$.

First we find $\frac{\partial u}{\partial x}$. Using the product rule we get

$$\frac{\partial u}{\partial x} = \frac{\partial}{\partial x} \left(x \cdot f(2x + y)\right) = \frac{\partial}{\partial x} (x) \cdot f(2x + y) + x \frac{\partial}{\partial x} \left(f(2x + y)\right)$$

Now we use the chain rule for a derivative of the composite function

$$\frac{\partial u}{\partial x} = f(2x + y) + x \frac{\partial f(v)}{\partial v} \frac{\partial}{\partial x} (2x + y) = f(2x + y) + 2x \frac{\partial f(v)}{\partial v}$$

where $v = 2x + y$

Then we find $\frac{\partial u}{\partial y}$ using the chain rule

$$\frac{\partial u}{\partial y} = \frac{\partial}{\partial y} \left(x \cdot f(2x + y)\right) = x \frac{\partial}{\partial y} \left(f(2x + y)\right) = x \frac{\partial f(v)}{\partial v} \frac{\partial}{\partial y} (2x + y) = x \frac{\partial f(v)}{\partial v}$$

Substituting $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$ into the original equation

$$x \frac{\partial u}{\partial x} - 2x \frac{\partial u}{\partial y} = u$$

gives

$$x \left(f(2x + y) + 2x \frac{\partial f(v)}{\partial v}\right) - 2x \left(x \frac{\partial f(v)}{\partial v}\right) = u(x, y)$$

or

$$x f(2x + y) + 2x^2 \frac{\partial f(v)}{\partial v} - 2x^2 \frac{\partial f(v)}{\partial v} = u \quad \Rightarrow \quad x f(2x + y) = u(x, y)$$

Thus, the function $u(x, y) = xf(2x + y)$ is a general solution of the equation

$$x \frac{\partial u}{\partial x} - 2x \frac{\partial u}{\partial y} = u$$

Answer: $u(x, y) = xf(2x + y)$ is a general solution of $x \frac{\partial u}{\partial x} - 2x \frac{\partial u}{\partial y} = u$.

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