y5y′−6y=0;y(0)=3,y′(0)=−7k2+5k−6=0→k1=−6,k2=1y=C1⋅e−6x+C2⋅ex1) y(0)=C1+C2=3→C2=3−C12) y′=−6C1⋅e−6x+C2⋅exy′(0)=−6C1+C2=−7−6C1+(3−C1)=−7−7C1=−10→C1=107,C2=3−107=1173) y=107⋅e−6x+117⋅ex

Check the solution to "y5y′−6y=0"

Name: y5y′−6y=0
Uploaded: 2022-02-23T17:22:57+00:00
Description: y5y′−6y=0

agreseza

Answered question

2021-12-30

y 5 y^{'} - 6 y = 0

Answer & Explanation

Melinda McCombs

Beginner2021-12-31Added 38 answers

Auxiliary equation of given differential equation is

m^{2} + 5 m - 6 = 0

Now roots of auxiliary equation

m^{2} + 5 m - 6 = 0

m^{2} + 6 m - m - 6 = 0

m (m + 6) - 1 (m + 6) = 0

(m + 6) (m - 1) = 0

m = - 6, 1

So, general solution of given differential equation is

y = c_{1} e^{- 6 x} + c_{2} e^{x}

Cheryl King

Beginner2022-01-01Added 36 answers

y 5 y^{'} - 6 y = 0; y (0) = 3, y^{'} (0) = - 7

k^{2} + 5 k - 6 = 0 \to k_{1} = - 6, k_{2} = 1

y = C_{1} \cdot e^{- 6 x} + C_{2} \cdot e^{x}

y (0) = C_{1} + C_{2} = 3 \to C_{2} = 3 - C_{1}

y^{'} = - 6 C_{1} \cdot e^{- 6 x} + C_{2} \cdot e^{x}

y^{'} (0) = - 6 C_{1} + C_{2} = - 7

- 6 C_{1} + (3 - C_{1}) = - 7

- 7 C_{1} = - 10 \to C_{1} = \frac{10}{7}, C_{2} = 3 - \frac{10}{7} = \frac{11}{7}

y = \frac{10}{7} \cdot e^{- 6 x} + \frac{11}{7} \cdot e^{x}

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