What is the constant #m#, if# y = e^(mx)# satisfies #(d^2y)/(dx^2) - 5dy/dx + 6y = 0# ?

3 Answers
Douglas K.
Feb 18, 2018

#m = 2# or #m= 3#

Explanation:

Given: #y = e^(mx)#

#dy/dx = me^(mx)#

#(d^2y)/(dx^2) = m^2e^(mx)#

Substituting into the equation #(d^2y)/(dx^2) - 5dy/dx + 6y = 0#:

#m^2e^(mx) - 5me^(mx) + 6e^(mx) = 0#

Dividing both sides by #e^(mx)#

#m^2 - 5m + 6 = 0#

Factor:

#(m -2)(m-3)=0#

#m = 2# or #m= 3#

Shiva Prakash M V
Feb 18, 2018

#m=3, m=2#

Explanation:

#y=e^(mx)#

#dy/dx=me^(mx)#

#(d^2y)/(dx^2)=m^2e^(mx)#

#"Differential equation is"#

#(d^2y)/(dx^2)-5(dy)/dx+6y=0#

#"Substituting"#

#m^2e^(mx)-5me^(mx)+6e^(mx)=0#

#(m^2-5m+6)e^(mx)=0#

#m^2-5m+6=0#

#m^2-2m-3m+6=0#

#m(m-3)-2(m-3)=0#

#(m-3)(m-2)=0#

#m-3=0, or m-2=0#

#m=3, m=2#

M. R.
Feb 18, 2018

m=3 or 2

Explanation:

#(dy)/dx=me^(mx)#

#(d^2y)/dx^2=m^2e^(mx)#

Substituting is equation,

# (d^2y)/dx^2 -5(dy)/dx +6y=0#

#m^2e^(mx) -5me^(mx)+6e^(mx)=0#

Dividing both sides by #e^(mx)#,

#m^2-5m+6=0#

#(m-3)(m-2)=0#

#m=3,2#

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