Given the general solution to #t^2y'' - 4ty' + 4y = 0# is #y= c_1t + c_2t^4#, how do I solve the problem #t^2y'' - 4ty' + 4y = -2t^2
, y(1) = 2, y'(1) =0#?

Question

Given the general solution to #t^2y'' - 4ty' + 4y = 0# is #y= c_1t + c_2t^4#, how do I solve the problem #t^2y'' - 4ty' + 4y = -2t^2
,   y(1) = 2,    y'(1) =0#?

Christopher Agresta · Accepted Answer

The answer is: #y=-t^4+t^2+2t#. If it is known a solution of the homogenous equation, now a particular solution of the non-homogenous equation has to be find. Since then in the second member there is #-2t^2#, for the similarity principle the particular solution is a square equation like this: #beta(t)=at^2+bt+c#. Now we calculate: #beta'(t)=2at+b#, and #beta''(t)=2a#. Now we have to find #a,b,c#. #beta(t)# has to satisfy the equation, so: #t^2(2a)-4t(2at+b)+4at^2+4bt+4c=-2t^2#
#2at^2-8at^2-4bt+4at^2+4bt+4c=-2t^2#
#-2at^2+4c=-2t^2# Then, for the principle of the identity between polynomials: #-2a=-2rArra=1#, and
#4c=0rArrc=0#.
#b# is not important, so we can assume that is 0. So: #beta(t)=t^2#. The solution of the non-homogeneous is: #y=c_1t+c_2t^4+t^2# and its derivative is: #y'=c_1+4c_2t^3+2t# Now we have to find #c_1# and #c_2# using the conditions: #y(1)=2# and #y'(1)=0#. #y'=c_1+4c_2t^3+2t#. So: #2=c_1+c_2+1rArrc_1+c_2=1# and #0=c_1+4c_2+2rArrc_1+4c_2=-2#. We can substitute #c_1=1-c_2# from the first equation to the second: #1-c_2+4c_2=-2rArr3c_2=-3rArrc_2=-1# And: #c_1=1-(-1)rArrc_1=2#. So, finally: #y=-t^4+t^2+2t#.

Ezra Adams · Answer

undefined To solve the problem $t^2y'' - 4ty' + 4y = -2t^2$, $y(1) = 2$, and $y'(1) = 0$, we can first find the general solution to the homogeneous differential equation $t^2y'' - 4ty' + 4y = 0$, which is given as $y = c_1t + c_2t^4$.

Next, we find the particular solution to the non-homogeneous equation $t^2y'' - 4ty' + 4y = -2t^2$. We can guess a particular solution of the form $y_p = At^2$, where $A$ is a constant. Substituting $y_p$ into the differential equation, we solve for $A$.

Once we have the particular solution $y_p$, we combine it with the general solution of the homogeneous equation to get the general solution of the non-homogeneous equation.

After obtaining the general solution, we apply the initial conditions $y(1) = 2$ and $y'(1) = 0$ to determine the values of the constants $c_1$ and $c_2$.

Given the general solution to #t^2y'' - 4ty' + 4y = 0# is #y= c_1t + c_2t^4#, how do I solve the problem #t^2y'' - 4ty' + 4y = -2t^2 ,&nbsp;&nbsp; y(1) = 2,&nbsp;&nbsp;&nbsp; y'(1) =0#?

Given the general solution to #t^2y'' - 4ty' + 4y = 0# is #y= c_1t + c_2t^4#, how do I solve the problem #t^2y'' - 4ty' + 4y = -2t^2 , y(1) = 2, y'(1) =0#?