Question

Solving a Differential Equation In Exercises \(1-10\) , solve the differential equation. $$ y^{\prime}=\sqrt{x} y $$

Short answer

The solution of the differential equation \(y' = \sqrt{x}y\) is \(y = ± Ae^{\frac{2}{3}x^{3/2}}\), where \(A\) is an arbitrary constant.

Step-by-step solution

Rewrite the Differential Equation in Separable Form

Rewrite the given differential equation \(y' = \sqrt{x}y\) in separable form. This can be done by dividing both side of the equation by \(y\) and multiplying both sides by \(dx\). This results in \(y'/y = \sqrt{x}\) => \(y'/y dx = \sqrt{x} dx\) => \(\frac{dy}{y} = \sqrt{x} dx\)

Integrate Both Sides of the Equation

Now perform the integration on each side of the equation. On the left side, the integral of \(1/y\) with respect to \(y\) is \(ln|y|\). On the right side, the integral of \(\sqrt{x}\) with respect to \(x\) is \(\frac{2}{3}x^{3/2}\). Therefore, we obtain \(ln|y| = \frac{2}{3}x^{3/2} + C\)

Solve for \(y\)

We can find \(y\) by exponentiating both sides of the equation to get rid of the natural logarithm \(ln\). This gives \(|y| = e^{\frac{2}{3}x^{3/2} + C}\). Considering the absolute values, this gives \(y = ± e^{C} e^{\frac{2}{3}x^{3/2}}\), which simplifies to \(y = ± Ae^{\frac{2}{3}x^{3/2}}\) where \(A\) = \(e^{C}\) is the arbitrary constant because \(C\) is an arbitrary constant.

Key concepts

Integration

Integration is a crucial step in solving separable differential equations. In the given problem, once we have rewritten the differential equation into a separable form, we can integrate both sides to gradually solve for the variable of interest.

• For the left side, we have the integral of \( \frac{1}{y} \) with respect to \( y \). This particular integral is known to simplify to \( \ln|y| \).
• For the right side, we handle the integral of \( \sqrt{x} \) with respect to \( x \). This is also expressed as \( x^{1/2} \), and its integration gives \( \frac{2}{3}x^{3/2} \).

Through these integrations, we bridge the formation of the differential equation to an equation that we can further manipulate. Notice that both integrals are elementary, but each serves its unique role in the solution process. It's essential to be comfortable with these integral formulas as they frequently appear in calculus.

Exponential Functions

The step involving exponential functions highlights their importance and how they interact with logarithms. Once we have the equation \( \ln|y| = \frac{2}{3}x^{3/2} + C \), converting from logarithmic to exponential form helps to directly solve for \( y \).

• This conversion uses the property that if \( \ln(a) = b \), then \( a = e^b \). Thus, \( |y| = e^{\frac{2}{3}x^{3/2} + C} \).
• Recognizing this relationship is key because it facilitates moving from the logarithmic expression back to a functional form.

When we exponentiate, the leftover constant \( C \) becomes part of the multiplicative constant in the solution. Exponential functions are powerful in this context because they can easily revert the logarithmic transformation, unlocking the expression we need for \( y \). They help link differential equations to exponential growth or decay scenarios, which are common applications.

Arbitrary Constants

Arbitrary constants appear frequently when solving differential equations, serving as crucial components for expressing the general solution.

• After integrating, we encounter the constant \( C \), which is arbitrary because indefinite integration allows for an infinite number of functions that differ by a constant.
• In our solution, \( e^C \) is rewritten as another constant \( A \) because \( e^C \) itself is arbitrary. Thus, the solution becomes \( y = \pm Ae^{\frac{2}{3}x^{3/2}} \).

Understanding these constants is vital as they offer flexibility. They mean that while the form of the expression remains constant, its specific shape or existence can vary depending on initial conditions or particular constraints applied to a problem. Arbitrary constants often allow us to adapt mathematical solutions to real-world problems, where the initial state might not be precisely known or fixed.