Solve the initial value problems in Exercises 67–70 for x as a function of t.
(t + 1) (dx/dt) = x² + 1 (for t > -1), x(0) = 0

검증된 단계별 안내

Rewrite the given differential equation \((t + 1) \frac{dx}{dt} = x^{2} + 1\) to isolate \(\frac{dx}{dt}\): \(\frac{dx}{dt} = \frac{x^{2} + 1}{t + 1}\).

Recognize that this is a separable differential equation. Rearrange terms to separate variables \(x\) and \(t\): \(\frac{dx}{x^{2} + 1} = \frac{dt}{t + 1}\).

Integrate both sides: \(\int \frac{dx}{x^{2} + 1} = \int \frac{dt}{t + 1}\).

Evaluate the integrals: The left integral is \(\arctan(x)\), and the right integral is \(\ln|t + 1| + C\), where \(C\) is the constant of integration. So, \(\arctan(x) = \ln|t + 1| + C\).

Use the initial condition \(x(0) = 0\) to find \(C\): Substitute \(t=0\) and \(x=0\) into the equation to solve for \(C\), then express \(x\) explicitly as a function of \(t\) by taking the tangent of both sides.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

영상 길이:

도움이 되었나요?

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Separable Differential Equations

A separable differential equation can be written so that all terms involving one variable are on one side and all terms involving the other variable are on the opposite side. This allows integration of both sides separately to find the solution. Recognizing and rearranging the given equation into separable form is essential for solving it.

Initial Value Problems (IVP)

An initial value problem specifies the value of the unknown function at a particular point, which helps determine the unique solution to a differential equation. Using the initial condition x(0) = 0 allows us to find the constant of integration after solving the differential equation.

Integration Techniques

Solving the separated differential equation requires integrating expressions involving variables and constants. Familiarity with standard integration methods, such as partial fractions or substitution, is necessary to evaluate the integrals and express x as a function of t.

Solve the initial value problems in Exercises 67–70 for x as a function of t.(t + 1) (dx/dt) = x² + 1 (for t > -1), x(0) = 0