Question

Find the general solution of the given higher-order differential equation. y''' − 3y'' − 4y' = 0

Answer 1

Higher order differential equation  $y^{III} -3y^{II} -4y^{I} =0$ has the general solution of $y=C_{1} e^{0} +C_{2} e^{-x} +C_{3} e^{4x}$.

Given higher order differential equation $y^{III} -3y^{II} -4y^{I} =0$.

We are required to find the general solution of the given higher order differential equation.

A differential equation is basically an equation which contains one or more terms and the derivatives of one variable with respect to the other variable dy/dx = f(x) Here “x” is basically an independent variable and “y” is  dependent variable.

Higher order differential equation $y^{III} -3y^{II} -4y^{I} =0$

Write the derivatives of y in powers of m.

$m^{2} -3m^{2} -4m=0$

m[$m^{2} -4m-4$]=0

m[$m^{2} -4m+m-4$]=0

m[m(m-4)+1(m-4)]=0

m(m-4)(m+1)=0

m=0,m=4,m=-1

Using the values of m in y=$y=C_{1} e^{m_{1}x } +C_{2} e^{m_{2}x } +C_{3} e^{m_{3} x}$.

y=$y=C_{1} e^{0x } +C_{2} e^{-1x } +C_{3} e^{4 x}$

$y=C_{1} e^{0} +C_{2} e^{-x} +C_{3} e^{4x}$

Hence the general solution of the higher order differential equation $y^{III} -3y^{II} -4y^{I} =0$ is $y=C_{1} e^{0} +C_{2} e^{-x} +C_{3} e^{4x}$.

Learn more about differential equations at brainly.com/question/18760518

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