Problem 26

Question

Solve the differential equation. $$ y^{\prime}=\ln x $$

Step-by-Step Solution

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Answer

The solution of the given differential equation $y' = \ln(x)$ is $y(x) = x \ln(x) - x + C$.

1Step 1: Understand the Type of The Differential Equation

The given equation $y' = \ln(x)$ is a first-order differential equation. There is only first derivative of the function $y$, and it equals to $\ln(x)$, a known function.

2Step 2: Applying the Integral

Since the derivative of $y$ is given, we can get $y(x)$ by integrating both sides. Integration of $y' = \ln(x)$ with respect to $x$ gives the equation $\int y' dx = \int \ln(x) dx$.

3Step 3: Evaluate the Integral

The integral of $\ln(x)$ is $x \ln(x) - x$ up to a constant of integration. Evaluating the integral, we get $y(x) = x \ln(x) - x + C$, where $C$ is the constant of integration.

Problem 26

Problem 27

Other exercises in this chapter

Problem 26

Find the integral. $$ \int \frac{\sqrt{1-x}}{\sqrt{x}} d x $$

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Problem 26

Use substitution to find the integral. $$ \int \frac{\sec ^{2} x}{\tan x(\tan x+1)} d x $$

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Problem 27

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{0}^{\infty} \cos \pi x d x $$

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Problem 27

Find the integral involving secant and tangent. $$ \int \tan ^{2} x \sec ^{2} x d x $$

View solution