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Question: How do I solve the differential equation \( y'' - 4y' - 5y = 0 \) with \( y( - 1) = 3 \) and \( y'( ...

How do I solve the differential equation y′′−4y′−5y=0y'' - 4y' - 5y = 0 with y(−1)=3y( - 1) = 3 and y′(−1)=9y'( - 1) = 9 ?

Explanation

Solution

Hint : To solve this question we should about basic of differential equation:
Differential equation: It is an equation that contains one or more functions with its derivative. The derivative of the function defines the rate of change of a function at a point.
For example: ay′′+by′+cy=0ay'' + by' + cy = 0 .

Complete step by step solution:
As the given differential equation is
y′′−4y′−5y=0,    y(−1)=3,    y′(−1)=9y'' - 4y' - 5y = 0,\,\,\,\,y( - 1) = 3,\,\,\,\,y'( - 1) = 9
We can assume y=emxy = {e^{mx}} be the auxiliary solution. Then we get
emx(m2−4m−5)=0{e^{mx}}({m^2} - 4m - 5) = 0
⇒m2−4m−5=0       [∵emx≠0,∀x]\Rightarrow {m^2} - 4m - 5 = 0\,\,\,\,\,\,\,[\because {e^{mx}} \ne 0,\forall x]
Find its roots,
⇒(m−5)(m+1)=0\Rightarrow (m - 5)(m + 1) = 0
So, (m−5)=0(m - 5) = 0
⇒m=5\Rightarrow m = 5
(m+1)=0(m + 1) = 0
⇒m=−1\Rightarrow m = - 1
We get,
⇒m=5,−1\Rightarrow m = 5, - 1
For two real roots m1≠m2{m_1} \ne {m_2} the general solution is of form y=c1em1x+c2em2xy = {c_1}{e^{{m_1}x}} + {c_2}{e^{{m_2}x}}
Thus the solution of the differential equation is
y(x)=c1e5x+c2e−xy(x) = {c_1}{e^{5x}} + {c_2}{e^{ - x}}
Now apply the first initial condition y(−1)=3y( - 1) = 3 given us
3=c1e−5+c2e3 = {c_1}{e^{ - 5}} + {c_2}e ………………. (1)(1)
Again apply second initial condition y′(−1)=9y'( - 1) = 9 given us
9=5c1e−5−c2e9 = 5{c_1}{e^{ - 5}} - {c_2}e …………… (2)(2)
Now adding equation (1)(1) and (2)(2) we get
3+9=c1e−5+5c1e−5+c2e−c2e3 + 9 = {c_1}{e^{ - 5}} + 5{c_1}{e^{ - 5}} + {c_2}e - {c_2}e
We solve further. We get,
12=6c1e−512 = 6{c_1}{e^{ - 5}}
⇒c1=2e5\Rightarrow {c_1} = 2{e^5}
Plugging the value c1=2e5{c_1} = 2{e^5} in equation (1)(1) we get
3=2+c2e3 = 2 + {c_2}e
⇒c2=e−1\Rightarrow {c_2} = {e^{ - 1}}
Plugging the value of c1{c_1} and c2{c_2} into the solution y=2e5e5x+e−1e−xy = 2{e^5}{e^{5x}} + {e^{ - 1}}{e^{ - x}} we get
y=2e5(1+x)+e−(1+x)=e1+x(2e5+1e)y = 2{e^{5(1 + x)}} + {e^{ - (1 + x)}} = {e^{1 + x}}(2{e^5} + \dfrac{1}{e}) Hence the required solution is
y=2e5(1+x)+e−(1+x)=e1+x(2e5+1e)y = 2{e^{5(1 + x)}} + {e^{ - (1 + x)}} = {e^{1 + x}}(2{e^5} + \dfrac{1}{e})
So, the correct answer is “ y=2e5(1+x)+e−(1+x)=e1+x(2e5+1e) y = 2{e^{5(1 + x)}} + {e^{ - (1 + x)}} = {e^{1 + x}}(2{e^5} + \dfrac{1}{e}) ”.

Note : The degree of the differential equation is the power of the highest order derivative, where the original equation is represented in the form of a polynomial equation in the derivative.
Order of derivative equation is the order of the highest order derivative present in the equation.
There is two methods to find the solution of differential equation:
I.Separation of variable
II. Integrating factor