Example 7 - Chapter 9 Class 12 Differential Equations
Last updated at April 16, 2024 by Teachoo
Variable separation - Equation given
Example 20
Misc 13 (MCQ)
Ex 9.3, 3
Ex 9.3, 8
Example 4
Ex 9.3, 7 Important
Example 5
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Ex 9.3, 2
Misc 4
Example 6
Ex 9.3, 5
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Example 7 Important You are here
Ex 9.3, 16
Ex 9.3, 12
Ex 9.3, 11 Important
Ex 9.3, 1 Important
Ex 9.3, 15 Important
Misc 7 Important
Ex 9.3, 4 Important
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Misc 6
Ex 9.3, 10 Important
Misc 12 Important
Misc 5 Important
Variable separation - Equation given
Last updated at April 16, 2024 by Teachoo
Example 7 Find the equation of the curve passing through the point (1 , 1) whose differential equation is 𝑥 𝑑𝑦= (2𝑥^2+1)𝑑𝑥(𝑥≠0) 𝑥 𝑑𝑦 = (2x2 + 1)dx dy = "(2x2 + 1)" /𝑥 dx dy = ("2x2" /𝑥+1/𝑥) dx dy = (𝟐𝒙+𝟏/𝒙) dx Integrating both sides. ∫1▒𝑑𝑦 = ∫1▒(2𝑥+1/𝑥) 𝑑𝑥 ∫1▒𝑑𝑦 = ∫1▒〖2𝑥 𝑑𝑥+〗 ∫1▒〖1/𝑥 𝑑𝑥〗 y = 2 𝑥2/2 + log |𝑥| + C y = 𝒙𝟐 + log |𝒙| + C Since the curve passes through point (1, 1) Putting x = 1, y = 1 is (1) 1 = 12 + log |𝟏| + C 1 = 1 + 0 + C 1 − 1 = C ∴ C = 0 Put C = 0 in (1) y = x2 + log |𝑥| + C y = x2 + log |𝒙| + 0 y = x2 + log |𝑥| Hence, the equation of curve is y = x2 + log |𝒙|