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Problem Statement
Find and plot the solution for $$ y''-10y'+25=0 \!$$.

Initial Conditions: $$ y(0)=1, y'(0)=0 \!$$

No excitation: $$ r(x)=0 \!$$

Background Information
Finding Homogeneous Solutions in cases where $$ \Delta=a^2-4b=0 \!$$

The two roots are a double root:

$$ \lambda_1=\lambda_2=\lambda=-a/2\!$$ (1)

The first homogeneous solution is:

$$ y_{h,1}(x)= e^{yx} \!$$ (2)

To find the 2nd homogeneous solution we use a trial solution:

$$ y_{h,2}=u(x)y_{h,1}(x) \!$$where u(x) is an unknown function of x (3)

$$ y'_{h,2}=uy'_{h,1}+u'y_{h,1} \!$$ (4)

$$ y_{h,2}=uy_{h,1}+2u'y'_{h,1}+u''y_{h,1}\!$$ (5)

Adding up equations (3), (4), and (5) we get:

$$ 0=u[y_{h,1}+ay'_{h,1}+by_{h,1}]+u'[2y'_{h,1}+ay_{h,1}]+uy_{h,1} \!$$ (6) We can simplify equation (6) by applying known equations:

$$ 0=u''y_{h,1} \!$$ (7)

Hence:

$$ u''=0 \!$$ (8)

Integrating with respect to x twice yields:

$$ u(x)=xc_3 \!$$ (9)

So $$ y_{h,2}=(x+c_3)y_{h,1}\!$$ (10)

The final homogeneous solution is:

$$ y_h=c_1y_{h,1}+c_2xy_{h,1} \!$$ (11)

Solution
$$ y''-10y'+25y=r(x) \!$$ (12)

The characteristic equation is:

$$ \lambda^2-10\lambda+25=0\!$$ (13)

The first homogeneous root is:

$$ \lambda=5 \!$$ (14)

Hence the homogeneous solution is:

$$ y_h(x)=c_1e^{5x}+c_2xe^{5x} \!$$ (15)

Apply the initial conditions y(0)=1 and y'(0)=0:

$$ y_h(0)= c_1 =1 \!$$ (16)

$$ y_h'(x)= 5c_1e^{5x}+5c_2xe^{5x}+c_2 \!$$<p style="text-align:right"> (17)

$$ y_h'(0)=5+c_2 \!$$<p style="text-align:right"> (18)

Hence:

$$c_1=1 \!$$<p style="text-align:right"> (19)

$$ c_2=-5 \!$$<p style="text-align:right"> (20)

The homogeneous equation is:

$$ y_h(x)=e^{5x}-5xe^{5x} \!$$<p style="text-align:right"> (21)

Since r(x)=0 the particular equation is zero. Hence the solution to the problem is:

$$ y(x)=e^{5x}-5xe^{5x} \!$$<p style="text-align:right"> (22)

The plot of the function is:



References:

Section 5 Notes