first order linear differential equation
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The document provides the steps to solve a first order linear differential equation of the form dy/dx + P(x)y = Q(x). It gives the standard form, explains how to find the integrating factor, and shows the work to solve the specific equation dy/dx + ysecx = tanx. The solution is found to be y = 1 - (x-c)/(secx+tanx).
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centuries, evolving its structure over time and space. In the present era, these changes have
accelerated due to factors such as agriculture and urbanization. Information regarding land use and
cover is essential for various planning and management tasks related to the Earth's surface,
providing crucial environmental data for scientific, resource management, policy purposes, and
diverse human activities.
Accurate understanding of land use and cover is imperative for the development planning
of any area. Consequently, a wide range of professionals, including earth system scientists, land
and water managers, and urban planners, are interested in obtaining data on land use and cover
changes, conversion trends, and other related patterns. The spatial dimensions of land use and
cover support policymakers and scientists in making well-informed decisions, as alterations in
these patterns indicate shifts in economic and social conditions. Monitoring such changes with the
help of Advanced technologies like Remote Sensing and Geographic Information Systems is
crucial for coordinated efforts across different administrative levels. Advanced technologies like
Remote Sensing and Geographic Information Systems
9
Changes in vegetation cover refer to variations in the distribution, composition, and overall
structure of plant communities across different temporal and spatial scales. These changes can
occur natural.
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first order linear differential equation
- 2. Solve the following first order linear differential equation. 𝒅𝒚 𝒅𝒙 + ysecx = tanx
- 3. Knowledge Standard Form: 𝒅𝒚 𝒅𝒙 + 𝑷 𝒙 . 𝒚 = 𝑸(𝒙) Integrating Factor: 𝒆 𝑷 𝒙 𝒅𝒙 Solution :: 𝒚. 𝑰. 𝑭 = 𝑸 𝒙 . 𝑰. 𝑭 𝒅𝒙 + 𝒄
- 4. Solution Given, P(x)= secx 𝒅𝒚 𝒅𝒙 + y. secx = tanx Q(x)= tanx # I.F = 𝐞 𝐬𝐞𝐜𝐱 𝐝𝐱 = 𝒆𝒍𝒏 𝒔𝒆𝒄𝒙+𝒕𝒂𝒏𝒙 = secx + tanx
- 5. ∴ The solution is 𝒚. 𝑰. 𝑭 = 𝑸 𝒙 . 𝑰. 𝑭 𝒅𝒙 + 𝑪 y(secx +tanx) = 𝒕𝒂𝒏𝒙. 𝒔𝒆𝒄𝒙 + 𝒕𝒂𝒏𝒙 𝒅𝒙 + 𝑪 = (𝒕𝒂𝒏𝒙. 𝒔𝒆𝒄𝒙 + 𝒕𝒂𝒏𝒙. 𝒕𝒂𝒏𝒙)𝒅𝒙 + 𝑪 = secx + tanx – x + C ∴ y = 1 – (x-c)/(secx+tanx) (Ans.)
- 6. Thank You