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NCERT Solutions for Class 12 Help.html Mathematics Part Ii Chapter 9 Differential Equations

Differential Equations Here is the CBSE Help.html Chapter 9 for Class 12 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 12 Help.html Differential Equations Chapter 9 NCERT Solutions for Class 12 Help.html Differential Equations Chapter 9 The following is a summary in Hindi and English for the academic year 2025-26. You can save these solutions to your computer or use the Class 12 Help.html.

Question 1
CBSEENMA12033080

At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (– 4, – 3). Find the equation of the curve given that it passes through ( 2, 1).

Solution
Let y = f(x) be equation of curve.
Now dy over dx is the slope of the tangent to the curve at the point (x, y)
From the given condition, 
          dy over dx space equals space 2 open parentheses fraction numerator negative 3 minus straight y over denominator negative 4 minus straight x end fraction close parentheses space space or space space space dy over dx space equals space 2 open parentheses fraction numerator straight y plus 3 over denominator straight x plus 4 end fraction close parentheses
Separating the variables,  we get,
                           fraction numerator 1 over denominator straight y plus 3 end fraction dy space equals space fraction numerator 2 over denominator straight x plus 4 end fraction dx
Integrating, integral fraction numerator 1 over denominator straight y plus 3 end fraction space dy space equals space 2 space integral fraction numerator 1 over denominator straight x plus 4 end fraction dx
therefore space space space log space open vertical bar straight y plus 3 close vertical bar space equals space 2 space log space open vertical bar straight x plus 4 close vertical bar plus straight c                                  ...(1)
Since curve passe through (-2, 1)
therefore space space space log space open vertical bar 1 plus 3 close vertical bar space equals space 2 space log space open vertical bar negative 2 plus 4 close vertical bar space plus space straight c
therefore space space log space 4 space equals space 2 space log space 2 space plus straight c space space space rightwards double arrow space space space space 2 space log space 2 space space equals space 2 space log space 2 plus space straight c space space rightwards double arrow space space straight c space equals space 0
therefore space space from space left parenthesis 1 right parenthesis comma space space log space open vertical bar straight y plus 3 close vertical bar space equals space 2 space log space open vertical bar straight x plus 4 close vertical bar
or space space log space open vertical bar straight y plus 3 close vertical bar space equals space log space open vertical bar straight x plus 4 close vertical bar squared
therefore space space space space open vertical bar straight y plus 3 close vertical bar space equals space open vertical bar straight x plus 4 close vertical bar squared space space space or space space space straight y plus 3 space equals space left parenthesis straight x plus 4 right parenthesis squared
which is required equation of curve.

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