Tricky Differential Equations Questions

Question

I am self studying differential equations, and ran into some problems I have never seen before.

The majority of problems I have solved have come int the form of $Ay''-by'+C = 0$. I clearly understand how to solve these, and arrive at the general solutions.

However, I am now encountering a problem like $2t^2y''-5ty'+6y = 0$ where they give a solution $y(t) = t^\frac32)$.

Does this now mean I have to work backwards from before since we already have a solution given to us?

Likewise, how does the general form look different in the case when $2t^2y'' - 5ty' + 6t = t^2 - 1$. I know this revolves around methods of undetermined coefficients, and have guessed $Ax^2+Bx+C$ and have differentiating it twice over for y'' and once over for y', but I still haven't organized my work together to arrive at a general solution.

Thank you

Answer 1

Hint: The differential equation $$2t^2y''-5ty'+6y = 0$$ or $$4t^2y''-10ty'+12y = 0$$ known as Euler differential equation, which can be solved with substitution $x=\ln2t$. In this case when the right side is zero, the differential equation $$(at)^2y''+paty'+qy = 0$$ is homogeneous and with $x=\ln at$ is converted to $$y''+\left(\dfrac{p}{a}-1\right)y'+\dfrac{p}{a^2}y = 0$$ then with characteristic equation $$\lambda^2+\left(\dfrac{p}{a}-1\right)\lambda+\dfrac{p}{a^2}=0$$ we may find it's general solution.

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Edit: After substitution $x=\ln 2t$, your equation is $$y''-\dfrac72y'+3y=2e^{2x}-2\tag{*}$$ from characteristic equation, the general solution is $$y_g=C_1e^{2x}+C_2e^{\frac32x}$$ and while the right side of equation (*) has item $e^{2x}$ which is one of the general solutions, then let $y=Axe^{2x}+B$ to find particular solution with undetermined coefficient method. After $A$ and $B$ were found, put $x=\ln 2t$ and return main independent variable. One may find further examples here.

Answer 2

This is an alternative route, but I think it is better to follow the answer by user 108128. Rewrite the differential equation as $$y''(t)-\frac{5}{2t}\,y'(t)+\frac{3}{t^2}\,y(t)=f(t)\,,$$ where $f$ is a source function. (Note that $f\equiv 0$ in the homogeneous case, and $$f(t)=\frac{t^2-1}{2t^2}=\frac{1}{2}\left(1-\frac{1}{t^2}\right)$$ for the nonhomogeneous equation in question.) Observe that $$\frac{\text{d}}{\text{d}t}\,\left(y'(t)-\frac{2}{t}\,y(t)\right)-\frac{1}{2t}\,\left(y'(t)-\frac{2}{t}\,y(t)\right)=f(t)\,.$$ Consequently, $$\frac{\text{d}}{\text{d}t}\,\frac{1}{\sqrt{t}}\,\left(y'(t)-\frac{2}{t}\,y(t)\right)=\frac{1}{\sqrt{t}}\,f(t)\,.$$ Ergo, $$y'(t)-\frac{2}{t}\,y(t)=\sqrt{t}\,F(t)\,,$$ where $\displaystyle F(t):=\int\,\frac{1}{\sqrt{t}}\,f(t)\,\text{d}t$. Finally, we write $$\frac{\text{d}}{\text{d}t}\,\left(\frac{1}{t^2}\,y(t)\right)=t^{-\frac{3}{2}}\,F(t)\,,$$ so that $$y(t)=t^2\,\int\,t^{-\frac{3}{2}}\,F(t)\,\text{d}t\,.$$

To solve the homogeneous solution (i.e., with $f\equiv 0$, we have $F(t)=-\frac{1}{2}\,A$ for some constant $A$. That is, $$y(t)=t^2\,\int\,\left(-\frac{1}{2}\,A\right)\,t^{-\frac{3}{2}}\,\text{d}t=t^2\,\left(A\,t^{-\frac{1}{2}}+B\right)=A\,t^{\frac{3}{2}}+B\,t^2\,,$$ for some constant $B$.
For the nonhomogeneous solution with $f(t)=\displaystyle\frac{1}{2}\,\left(1-\frac{1}{t^2}\right)$, we get $$F(t)=\int\,\frac{1}{2}\left(t^{-\frac{1}{2}}-t^{-\frac{5}{2}}\right)\,\text{d}t=t^{\frac{1}{2}}+\frac{1}{3}\,t^{-\frac{3}{2}}-\frac12\,a\,,$$ where $a$ is a constant. That is, $$y(t)=t^2\,\int\,t^{-\frac{3}{2}}\,\left(t^{\frac{1}{2}}+\frac{1}{3}\,t^{-\frac{3}{2}}-\frac12\,a\right)\,\text{d}t=t^2\,\left(\ln(t)-\frac{1}{6}t^{-2}+a\,t^{-\frac{1}{2}}+b\right)\,,$$ for some constant $b$. Thus, $$y(t)=t^2\,\ln(t)-\frac{1}{6}+a\,t^{\frac{3}{2}}+b\,t^2\,.$$

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P.S. You can also rewrite the differential equation as $$\frac{\text{d}}{\text{d}t}\,\left(y'(t)-\frac{3}{2t}\,y(t)\right)-\frac{1}{t}\,\left(y'(t)-\frac{3}{2t}\,y(t)\right)=f(t)\,.$$ This leads to $$y(t)=t^{\frac32}\,\int\,t^{-\frac{1}{2}}\,\tilde{F}(t)\,\text{d}t\,,$$ where $$\tilde{F}(t):=\int\,\frac{1}{t}\,f(t)\,\text{d}t\,.$$

Answer 3

I think that it's informative to look at degrees, considering $t$ as raising the degree with one and $\frac{d}{dt}$ as lowering the degree with one. For all the terms in your differential equation you have as many $t$ factors as applications of $\frac{d}{dt}$ and the degree therefore remains unchanged. Even more important is that the degree changes in the same way for every term. This means that if the solution in some way can be written as a sum $y(t) = \sum_{\alpha \in A} c_\alpha t^\alpha$ then the differential equation works on each term separately. In this case you get $2t^2 y'' - 5ty' + 6y = \sum_{\alpha \in A} (2\alpha(\alpha-1)-5\alpha+6) c_\alpha t^\alpha.$ so for the differential equation to be satisfied you must have $(2\alpha(\alpha-1)-5\alpha+6) c_\alpha = 0,$ i.e. $2(\alpha-2)(\alpha-\frac32)c_\alpha=0.$ Thus $c_\alpha$ can only be nonzero for $\alpha=2$ and $\alpha=\frac32.$