{"id":33029,"date":"2026-08-31T15:35:26","date_gmt":"2026-08-31T15:35:26","guid":{"rendered":"https:\/\/www.vedprep.com\/exams\/?p=33029"},"modified":"2026-08-31T15:35:26","modified_gmt":"2026-08-31T15:35:26","slug":"charpit-s-method-4","status":"publish","type":"post","link":"https:\/\/www.vedprep.com\/exams\/upsc\/charpit-s-method-4\/","title":{"rendered":"Charpit\u2019s Method: 5 Proven Steps to Master First Order PDEs"},"content":{"rendered":"
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Charpit\u2019s Method: 5 Proven Steps to Master First Order PDEs for UPSC<\/h1>\n

The Charpit\u2019s method<\/strong> is a powerful technique for solving first-order partial differential equations (PDEs) that every UPSC aspirant should master for Mathematics and Physics optional papers. This method transforms complex PDEs into solvable ordinary differential equations (ODEs), making it indispensable for exams like CSIR NET, IIT JAM, and GATE. By following a systematic approach, candidates can solve these equations efficiently and accurately.<\/p>\n

Charpit\u2019s Method: Key Concepts<\/h2>\n

Developed by French mathematician Ludovic Charpit in the 1840s, Charpit\u2019s method<\/strong> is specifically designed to tackle first-order PDEs, which involve unknown multivariable functions and their first partial derivatives. The method works by converting the PDE into a system of characteristic ODEs, simplifying the problem-solving process. This technique is widely used in physics for wave propagation, engineering for stress analysis, and economics for optimal control models.<\/p>\n

For UPSC aspirants, mastering Charpit\u2019s method<\/strong> can be a game-changer. It allows you to break down complex problems into manageable steps, ensuring you can solve them within the limited timeframe of the exam. Whether you’re preparing for CSIR NET, IIT JAM, or GATE, understanding and applying this method will significantly boost your performance in the analytical sections of your optional subjects.<\/p>\n

The Core Mechanics of Charpit\u2019s method<\/strong> Explained<\/h2>\n

To apply Charpit\u2019s method<\/strong>, follow these five critical steps:<\/p>\n

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  1. Rewrite the PDE:<\/strong> Express the PDE in the form F(x, y, u, p, q) = 0<\/code>, where p = \u2202u\/\u2202x<\/code> and q = \u2202u\/\u2202y<\/code>. This step organizes the variables needed for the subsequent characteristic equations.<\/li>\n
  2. Form the characteristic ODEs:<\/strong> Differentiate F<\/code> with respect to p<\/code> and q<\/code> to get F_p<\/code> and F_q<\/code>. The characteristic equations are then given by:<\/li>\n<\/ol>\n

    dx \/ F_p = dy \/ F_q = du \/ (p F_p + q F_q)<\/code><\/p>\n

      \n
    1. Integrate the ODEs:<\/strong> Integrate the first two ratios to obtain a first invariant I\u2081(x, y, u) = C\u2081<\/code>. Integrate the third ratio with one of the previous ones to get a second invariant I\u2082(x, y, u) = C\u2082<\/code>.<\/li>\n
    2. Express the general solution:<\/strong> Combine the invariants to express the general solution implicitly as \u03a6(I\u2081, I\u2082) = 0<\/code>, where \u03a6<\/code> is an arbitrary function determined by boundary conditions.<\/li>\n
    3. Verify the solution:<\/strong> Substitute the obtained solution back into the original PDE to ensure it holds true. This step is crucial to avoid errors and ensure accuracy.<\/li>\n<\/ol>\n

      For competitive exams, it’s essential to memorize these steps and practice with various examples to reinforce understanding.<\/p>\n

      Step-by-Step Guide: Solving a First Order PDE Using Charpit\u2019s method<\/strong><\/h2>\n

      Let\u2019s take a closer look at how to apply Charpit\u2019s method<\/strong> with a practical example. Consider solving the PDE:<\/p>\n

      u_x^2 + u_y^2 = 4u<\/code><\/p>\n

      Step 1: Rewrite the PDE<\/strong><\/p>\n

      Introduce the notations p = \u2202u\/\u2202x<\/code> and q = \u2202u\/\u2202y<\/code>. Define F(p, q, u) = p^2 + q^2 - 4u = 0<\/code>.<\/p>\n

      Step 2: Form the characteristic ODEs<\/strong><\/p>\n

      The characteristic equations are derived as:<\/p>\n

      dx \/ F_p = dy \/ F_q = du \/ (p F_p + q F_q)<\/code><\/p>\n

      Here, F_p = 2p<\/code>, F_q = 2q<\/code>, and p F_p + q F_q = 2(p^2 + q^2) = 8u<\/code>. Thus, the characteristic system becomes:<\/p>\n

      dx \/ 2p = dy \/ 2q = du \/ 8u<\/code><\/p>\n

      Step 3: Integrate the ODEs<\/strong><\/p>\n

      From the first two ratios, we get:<\/p>\n

      dx \/ p = dy \/ q<\/code><\/p>\n

      Integrating this gives the first invariant:<\/p>\n

      x - y = C_1<\/code><\/p>\n

      Using the third ratio with dx \/ 2p = du \/ 8u<\/code>, we get:<\/p>\n

      4u dx = p du<\/code><\/p>\n

      Substituting p = du\/dx<\/code> leads to:<\/p>\n

      4u dx = (du\/dx) du<\/code><\/p>\n

      Integrating this produces:<\/p>\n

      u = (x^2 + y^2)\/4 + C_2<\/code><\/p>\n

      Step 4: Express the general solution<\/strong><\/p>\n

      The complete integral of the original PDE is:<\/p>\n

      u(x,y) = (x^2 + y^2)\/4 + C<\/code><\/p>\n

      Step 5: Verify the solution<\/strong><\/p>\n

      Verify by substituting back into the original PDE:<\/p>\n

      u_x = x\/2, u_y = y\/2<\/code><\/p>\n

      Thus, u_x^2 + u_y^2 = (x^2 + y^2)\/4<\/code>. Since 4u = x^2 + y^2 + 4C<\/code>, the equality holds true for any constant C<\/code>.<\/p>\n

      Common Mistakes to Avoid with Charpit\u2019s method<\/strong><\/h2>\n

      Many candidates make critical errors when applying Charpit\u2019s method<\/strong>. Here are some common pitfalls and how to avoid them:<\/p>\n