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

Last updated at Dec. 11, 2019 by Teachoo

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Example 4 Deleted for CBSE Board 2023 Exams

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Example 8 Deleted for CBSE Board 2023 Exams

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Example 25 Deleted for CBSE Board 2023 Exams

Example 26 You are here

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Chapter 9 Class 12 Differential Equations

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Last updated at Dec. 11, 2019 by Teachoo

Example 26 Find the particular solution of the differential equation log(ππ¦/ππ₯)=3π₯+4π¦ given that π¦=0 π€βππ π₯=0 log(ππ¦/ππ₯)=3π₯+4π¦ ππ¦/ππ₯ = e(3x + 4y) ππ¦/ππ₯ = e3x e4y Separating the variables ππ¦/π^4π¦ = e3x dx eβ4y dy = e3x dx Integrating both sides β«1βγπ^(β4π¦) ππ¦γ=β«1βπ^3π₯ ππ₯ π^(β4π¦)/(β4)=π^3π₯/3+πΆ 0=π^3π₯/3+π^(β4π¦)/4+πΆ π^3π₯/3 + π^(β4π¦)/4 + C = 0 (4π3π₯" + " 3π^(β4π¦))/12 + C = 0 (4π3π₯" + " 3π^(β4π¦) + 12πΆ)/12 " = 0" 4π^3π₯ + 3π^(β4π¦) + 12C = 0 Given y = 0 when x = 0 β¦(1) Putting x = 0 & y = 0 in (1) 4π^(3(0)) + 3π^(β4(0)) + 12C = 0 4π^0 + 3π^0 + 12C = 0 4 + 3 + 12C = 0 7 + 12C = 0 12C = β 7 C = (β7)/12 Putting value of C in (1) 4π^3π₯ + 3π^(β4π¦) + 12C = 0 4π^3π₯ + 3π^(β4π¦) + 12((β7)/12) = 0 ππ^ππ + ππ^(βππ) β 7 = 0