Verifying that $e^t$ is a solution to $(2t − t^2)y''+ (t^2-2)y'+2(1-t)y=0$, then deriving the general solution.

Question

Verify that $e^t$ is a solution of the differential equation

$$(2t − t^2)y''+ (t^2-2)y'+2(1-t)y=0.$$

Use this solution to find the general solution.

Answer 1

$$(2t-t^2)y''+(t^2-2)y'+2(1-t)y=0$$

First verify that $e^t$ is a solution:

$$(2t-t^2)e^t+(t^2-2)e^t+2(1-t)e^t=(2t-t^2+t^2-2+2-2t)e^t=0e^t=0$$

Also note that based on the format of the equation, there is likely a power function that works. So try $y=t^n$.

$$n(n-1)(2t-t^2)t^{n-2}+n(t^2-2)t^{n-1}+2(1-t)t^n=0$$ $$\implies(n-2)t^{n+1}-(n^2-n-2)t^n+(2n^2-4n)t^{n-1}=0$$ $$\implies(n-2)(t^{n+1}-(n+1)t^n+2nt^{n-1})=0$$ $$\implies n=2$$

So $y=t^2$ is a solution. Since any multiple of a solution of this DE is another solution, all linear combinations work.

Note that we could also find the $y=t^2$ solution by simply noticing $2t-t^2+t(t^2-2)=t^3-t^2=t^2(1-t)$, which gives it away.

Answer: $$y=Ae^t+Bt^2$$

Answer 2

It is easy to verify that $y_1(t)=e^t$ is a solution to your differential equation, just by substituting it into your ODE, as shown in @RomanChokler's answer.

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Since we know one solution, and we are trying to find a second linearly independent one, it is ideal to use reduction of order. We will assume the other solution to be of the form $y_2(t)=v(t)y_1(t)$, where $v(t)$ is an unknown function to be determined. Hence, in our case, we will substitute $y_2(t)=v(t)e^t$ into our ODE. $$(2t-t^2)\color{green}{y''}+(t^2-2)\color{#000090}{y'}+2(1-t)\color{#aa0000}{y}=0 \tag{1}$$ $$(2t-t^2)\color{green}{(e^t\cdot v''+2e^t\cdot v'+e^t\cdot v)}+(t^2-2)\color{#000090}{(e^t\cdot v'+e^t\cdot v)}+2(1-t)\color{#aa0000}{(e^t\cdot v)}=0$$ Gathering derivatives of $v$ and simplifying, we see that we must solve the following second order differential equation: $$t(2-t)v''+(-t^2+4t-2)v'=0 \tag{2}$$ Solving the above is not difficult: notice that the term involving the $v$ has dropped out. Thus, it is evident that we can reduce the order of $(2)$ with the change of variable $w=v'$, giving a separable differential equation: $$w'=\frac{-t^2+4t-2}{t(t-2)}w \tag{3}$$ Using the fact that $\dfrac{-t^2+4t-2}{t(t-2)}=\dfrac{1}{t}+\dfrac{1}{t-2}-1$, it can be shown with ease that the solution to $(3)$ is: $$w(t)=\frac{c_1 t(t-2)}{e^t}$$ Now recall our change of variable, and integrate both sides with respect to $t$. We obtain the following solution via integration by parts: $$v(t)=\frac{k_1 t^2}{e^t}+k_2 \tag{4}$$ Where $k_1:=-c_1$. Since we now have $v(t)$, we conclude that the second linearly independent solution is $y_2(t)=t^2$.