Question

Use the method outlined in Problem 28 to solve the given differential equation. $$ t^{2} y^{\prime \prime}-2 t y^{\prime}+2 y=4 t^{2}, \quad t>0 ; \quad y_{1}(t)=t $$

Short answer

Answer: The general solution for the given differential equation is \(y(t) = c_1t -\frac{c_2}{2t^2} - 2t^2 + \frac{t^4}{4}\)

Step-by-step solution

Identify the complementary function

The given differential equation is $$t^2 y'' - 2ty' + 2y = 4t^2$$ with the first complementary function \(y_1(t) = t\). Since we have one complementary function, we can find the Wronskian \(W(t)\) which is given by $$W(t) = y_1(t)y_1'(t) = t(1) = t$$ Now, let's calculate the particular integral.

Find the particular integral

To find the particular integral \(y_p(t)\), we use variation of parameters method as follows: First, we find the second complementary function \(y_2(t)\) using the formula: $$y_2(t)=y_1(t)\int{\frac{1}{y_1^2(t)W(t)} \ dt}$$ Substituting the values, we get $$y_2(t)=t\int{\frac{1}{t^2 \cdot t} \ dt} = t\int{\frac{1}{t^3} \ dt}$$ Integrating with respect to \(t\), we get $$y_2(t)= -\frac{1}{2t^2}$$ Second, we find the particular integral \(y_p(t)\) using the formula: $$y_p(t)= -y_1(t)\int{y_2(t) Q(t) \ dt}+y_2(t)\int{y_1(t) Q(t) \ dt}$$ Here, \(Q(t)\) is the right-hand side of the given non-homogeneous differential equation, which is \(4t^2\). Substituting the values, we get $$y_p(t)=-t\int{(-\frac{1}{2t^2})(4t^2) \ dt}+(-\frac{1}{2t^2})\int{t(4t^2) \ dt}$$ Simplifying and integrating, we get $$y_p(t)=-t\int{2 \ dt}+(-\frac{1}{2t^2})\int{4t^3 \ dt}$$ $$y_p(t)=-2t^2+\frac{t^4}{4}$$

Obtain the general solution

Now that we have the complementary function and the particular integral, we can obtain the general solution for the given differential equation as follows: $$y(t) = c_1y_1(t) + c_2y_2(t) + y_p(t)$$ Substituting the values, $$y(t) = c_1t + c_2(-\frac{1}{2t^2}) + (-2t^2 + \frac{t^4}{4})$$ Thus, the general solution for the given differential equation is $$y(t) = c_1t -\frac{c_2}{2t^2} - 2t^2 + \frac{t^4}{4}$$