Complete Question
Find a particular solution to [tex]ty''-(1+t)y'+y=t^2e^t , t > 0[/tex]
[tex]y_1 = 1 + t , \ y_2 = e^t[/tex]
Answer:
The solution is [tex]y_p = -e^{2t} + t + \frac{3t^2}{2} + \frac{t^3}{2}[/tex]
Step-by-step explanation:
Now the given equation is [tex]ty''-(1+t)y'+y=t^2e^t , t > 0[/tex]
dividing through by t
We have
[tex]y'' - \frac{1 + t }{t}y' + \frac{1}{t} y = te^t \\\\ let's\ call\ it \ g(t)[/tex]
Now for the given initial value we can generate our Wronskian as follows
[tex]W(y_1 ,y_2) = | \left y_1} \atop {y_1'}} \right. \left y_2} \atop {y_2'}} \right | = | \left {(1 + t)} \atop {1}} \right. \left e^t} \atop {e^t}} \right | = (1 + t )e^t - e^t = te^t[/tex]
Now applying method of variation of parameters to obtain the particular solution
So here we assume that
[tex]y_p = v_1 y_1 + v_2 y_2[/tex]
So
[tex]v_1 = - \int\limits {\frac{y_2 (t) g(t)}{W(t)} } \, dt = - \int\limits {\frac{e^t te^t}{te^t} } = - \int\limits { e^t} } = -e^t[/tex]
And
[tex]v_2 = - \int\limits {\frac{y_1 (t) g(t)}{W(t)} } \, dt = - \int\limits {\frac{(1 + t) te^t}{te^t} } = - \int\limits (1 + t) = t + \frac{t^2}{2}[/tex]
So
[tex]y_p = -e^t * e^t + t + \frac{t^2}{2} * 1 + t[/tex]
[tex]y_p = -e^{2t} + t + \frac{3t^2}{2} + \frac{t^3}{2}[/tex]
