1. Chapter 9 Class 12 Differential Equations (Term 2)
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Example 26 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Dec. 11, 2019 by

Example 26 - Find particular solution log (dy/dx) =3x + 4y

Example 26 - Chapter 9 Class 12 Differential Equations - Part 2
Example 26 - Chapter 9 Class 12 Differential Equations - Part 3

  1. Chapter 9 Class 12 Differential Equations (Term 2)
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    3. Examples

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Transcript

Example 26 Find the particular solution of the differential equation log(๐‘‘๐‘ฆ/๐‘‘๐‘ฅ)=3๐‘ฅ+4๐‘ฆ given that ๐‘ฆ=0 ๐‘คโ„Ž๐‘’๐‘› ๐‘ฅ=0 log(๐‘‘๐‘ฆ/๐‘‘๐‘ฅ)=3๐‘ฅ+4๐‘ฆ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = e(3x + 4y) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = e3x e4y Separating the variables ๐‘‘๐‘ฆ/๐‘’^4๐‘ฆ = e3x dx eโˆ’4y dy = e3x dx Integrating both sides โˆซ1โ–’ใ€–๐‘’^(โˆ’4๐‘ฆ) ๐‘‘๐‘ฆใ€—=โˆซ1โ–’๐‘’^3๐‘ฅ ๐‘‘๐‘ฅ ๐‘’^(โˆ’4๐‘ฆ)/(โˆ’4)=๐‘’^3๐‘ฅ/3+๐ถ 0=๐‘’^3๐‘ฅ/3+๐‘’^(โˆ’4๐‘ฆ)/4+๐ถ ๐‘’^3๐‘ฅ/3 + ๐‘’^(โˆ’4๐‘ฆ)/4 + C = 0 (4๐‘’3๐‘ฅ" + " 3๐‘’^(โˆ’4๐‘ฆ))/12 + C = 0 (4๐‘’3๐‘ฅ" + " 3๐‘’^(โˆ’4๐‘ฆ) + 12๐ถ)/12 " = 0" 4๐‘’^3๐‘ฅ + 3๐‘’^(โˆ’4๐‘ฆ) + 12C = 0 Given y = 0 when x = 0 โ€ฆ(1) Putting x = 0 & y = 0 in (1) 4๐‘’^(3(0)) + 3๐‘’^(โˆ’4(0)) + 12C = 0 4๐‘’^0 + 3๐‘’^0 + 12C = 0 4 + 3 + 12C = 0 7 + 12C = 0 12C = โ€“ 7 C = (โˆ’7)/12 Putting value of C in (1) 4๐‘’^3๐‘ฅ + 3๐‘’^(โˆ’4๐‘ฆ) + 12C = 0 4๐‘’^3๐‘ฅ + 3๐‘’^(โˆ’4๐‘ฆ) + 12((โˆ’7)/12) = 0 ๐Ÿ’๐’†^๐Ÿ‘๐’™ + ๐Ÿ‘๐’†^(โˆ’๐Ÿ’๐’š) โˆ’ 7 = 0

Chapter 9 Class 12 Differential Equations (Term 2)
Serial order wise

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