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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
9장, 문제 9.4.40

39–42. Special equations A special class of first-order linear equations have the form a(t)y'(t)+a'(t)y(t)=f(t), where a and f are given functions of t. Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form
a(t)y'(t) + a'(t)y(t) = d/dt (a(t)y(t)) = f(t). 
Therefore, the equation can be solved by integrating both sides with respect to t. Use this idea to solve the following initial value problems. 


t³y′(t) + 3t²y = (1 + t)/t, y(1) = 6

검증된 단계별 안내
1
Recognize that the given differential equation is of the form \(a(t)y'(t) + a'(t)y(t) = f(t)\), where \(a(t) = t^3\). This means the left side can be expressed as the derivative of the product \(a(t)y(t)\), i.e., \(\frac{d}{dt}(t^3 y(t))\).
Rewrite the equation using this product rule form: \(\frac{d}{dt}(t^3 y(t)) = \frac{1 + t}{t}\).
Integrate both sides with respect to \(t\) to find \(t^3 y(t)\): \(\int \frac{d}{dt}(t^3 y(t)) \, dt = \int \frac{1 + t}{t} \, dt\).
Simplify the integral on the right side by splitting the fraction: \(\int \left( \frac{1}{t} + 1 \right) dt = \int \frac{1}{t} dt + \int 1 dt\).
After integrating, solve for \(y(t)\) by dividing both sides by \(t^3\), then use the initial condition \(y(1) = 6\) to find the constant of integration.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
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주요 개념

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

First-Order Linear Differential Equations

These are differential equations of the form y' + p(t)y = q(t), where p and q are functions of t. They can often be solved using integrating factors or by recognizing patterns that simplify the equation. Understanding their structure is essential for applying appropriate solution methods.
추천 영상:
07:39
Classifying Differential Equations

Product Rule and Recognizing Derivatives of Products

The product rule states that d/dt [a(t)y(t)] = a(t)y'(t) + a'(t)y(t). Recognizing when the left side of a differential equation matches this derivative allows rewriting the equation in a simpler form, facilitating direct integration to find solutions.
추천 영상:
05:18
The Product Rule

Initial Value Problems and Integration

An initial value problem specifies the value of the solution at a particular point, enabling determination of the integration constant after solving the differential equation. Integrating both sides with respect to t and applying the initial condition yields the unique solution.
추천 영상:
05:03
Initial Value Problems