Solving homogeneous differential equation
Solving homogeneous differential equation
Last updated at August 13, 2026 by Teachoo
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Ex 9.4, 17 Which of the following is a homogeneous differential equation ? (A) (4š„+6š¦+5)šš¦ā(3š¦+2š„+4)šš„=0 (B) (š„š¦)šš„ā(š„^3+š¦^3 )šš¦=0 (C) (š„^3+2š¦^2 )šš„+2š„š¦ šš¦=0 (D) š¦^2 šš„+(š„^2+š„š¦āš¦^2 )šš¦=0Let us check each equation one by one Checking Option (A) Differential equation can be written as (4š„+6š¦+5)šš¦ā(3š¦+2š„+4)šš„ šš¦/šš„ = ((3š¦ + 2š„ + 4))/((4š„ + 6š¦ + 5)) Let F(x, y) = šš¦/šš„ = ((3š¦ + 2š„ + 4))/((4š„ + 6š¦ + 5)) Finding F(šx, šy) F(šx, šy) = (2šš„ + 3šš¦ + 4)/(4šš„ + 6šš¦ + 5) ā šĀ° F(x, y) ā“ The given equation is not homogenous Checking Option (B) (B) Differential equation can be written as (š„š¦)šš„ā(š„^3+š¦^3 )šš¦ = 0 šš¦/šš„ = š„š¦/(š„^3 + š¦^3 ) Let F(x, y) = šš¦/šš„ = š„š¦/(š„^3 + š¦^3 ) Finding F(šx, šy) F(šx, šy) = (šš„ šš¦)/(š^3 š„^3 + š^3 š¦^3 ) = (š^2 š„š¦)/(š^3 [š„^3 + š¦^3 ] ) = š„š¦/š(š„^3+š¦^3 ) ā šĀ° F(x, y) ā“ The given equation is not homogenous Checking Option (C) (š„^3+2š¦^2 )šš„+2š„š¦ šš¦=0 (x3 + 2y2) dx = ā2xy dy šš¦/šš„ = (ā(š„^3 + 2š¦^2))/2š„š¦ Let F(x, y) = šš¦/šš„ = (ā(š„^3 + 2š¦^2))/2š„š¦ Finding F(šx, šy) F(šx, šy) = (ā(š^3 š„^3 + 2š^2 š¦^2))/2šš„šš¦ = (āć6š„ć^3 + 2š¦^2)/2š„š¦ ā šĀ° F(x, y) ā“ The given equation is not homogenous Checking Option (D) y2 dx + (x2 ā xy ā y2) dy = 0 y2 dx = (x2 ā xy ā y2)dy šš¦/šš„ = š¦^2/(š„^2 ā š„š¦ ā š¦2) Let F(x, y) = šš¦/šš„ = š¦^2/(š„^2 ā š„š¦ ā š¦2) Finding F(šx, šy) F(šx, šy) = ćāš^(2 ) š¦ć^2/(š^(2 ) (š„^2 ā š„š¦ ā š¦2)) = š¦^2/(š„^2 ā š„š¦ ā š¦2) = šĀ°F (x, y) F (x, y) is š homogenous function of degree zero. ā“ Given equation is a homogenous differential equation. Hence, (D) is the correct answer.