Answer to Question #231189 in Differential Equations for Randal Rodriguez

Question

eliminate the arbitrary constant using any method

14.y\:=\:−2C_1e^{−3x}\:+\:C_2e^{4x}

Answer 1

Solution

If y(x) = −2C₁e^−3x + C₂e^4x  <=eq.1

then

y’(x) = 6C₁e^−3x + 4C₂e^4x  <= eq.2

y’’(x) = -18C₁e^−3x + 16C₂e^4x  <= eq.3

From 3* eq.1+ eq.2 => y’(x)+3 y(x) = 7 C₂e^4x     <= eq.4

From 3* eq.2+ eq.2 => 3 y’(x)+y’’(x) = 28 C₂e^4x     <= eq.5

From 4* eq.4-eq.5 => 4y’(x)+12y(x) - 3y’(x) - y’’(x) = 0  =>  y’’(x) – y’(x) – 12y(x) = 0

So the answer is y’’(x) – y’(x) – 12y(x) = 0