Chapter 2: Problem 4

The initial value problem $$ \frac{d y}{d x}=\frac{y^{2}+x y}{x^{2}}, \quad y(1)=1 $$ does not fall into the class of problems considered in this chapter. However, if one substitutes $y(x)=x z(x)$ into the differential equation, one obtains an equation for $z(x)$ that can be solved. Use this substitution to solve the initial value problem for $y(x)$.

Short Answer

Expert verified

The solution to the initial value problem is $y(x) = \frac{x^2}{1 +x}$.

Step by step solution

Substitute Given Function

We start by substituting $y(x) = xz(x)$ into the given differential equation. Taking derivative of both sides gives $\frac{d y}{d x} = z(x) + x \frac{d z}{d x}$. We substitute these into the original differential equation, obtaining $\frac{d y}{d x} = z(x) + x \frac{d z}{d x} = \frac{(x z(x))^2+ x (x z(x))}{x^2}$. Simplifying, we find $z + x \frac{d z}{d x} = z^2 + z$.

Rearrange Equation

We now rearrange the differential equation to isolate the derivative on one side. This gives us $\frac{d z}{d x} = z(z - 1)$.

Solve Separable Differential Equation

This equation is now separable, meaning we can rearrange it as a product of functions of z and x and then integrate. $ \frac{d z}{z(z - 1)} = \frac{d x}{x} $. Integrate both sides as $\int \frac{d z}{z(z - 1)} = \int \frac{d x}{x}$ to get $ \ln|x| = - \ln|z - 1| + \ln|z|$. This simplifies to $ \ln|x| = \ln \left|\frac{z}{z-1}\right|$. Taking exponential of both sides, we get $|x| = \left|\frac{z}{z-1}\right|$. Since we started with y(1) = 1, we know that z(1) = 1 and therefore, we can drop the absolute value on the right side but keep it on the left side.

Rearrange and Substitute Back for y

We rearrange the result to solve for z: $z = \frac{x}{1 \pm x}$. Because the initial condition is y(1) = 1, we choose the upper sign, so $z = \frac{x}{1 +x}$. Finally, remembering our substitution y(x) = xz(x), this gives us the solution for our original differential equation: $y(x) = x \cdot \frac{x}{1 +x} = \frac{x^2}{1 +x}$.