Question

Consider the following differential equation:
\[ t^2\dfrac{d^2y}{dt^2}+7t\dfrac{dy}{dt}+8ty=10\sin(t) \]
Which one of the following options is correct?

• A. It is a linear differential equation
• B. It is a nonlinear differential equation
• C. It is a time-invariant differential equation
• D. It is a second-order partial differential equation

Hint:

Check if y and its derivatives appear only to the first power for linearity, separately from whether the coefficients depend on t for time invariance.

Answer 1

The Correct Option is A

We can also test linearity and time invariance formally, using the operator that the equation defines, rather than just inspecting the terms.

Define the operator $L[y]=t^2y''+7ty'+8ty$. To test linearity, check whether $L[c_1y_1+c_2y_2]=c_1L[y_1]+c_2L[y_2]$ for any constants $c_1,c_2$ and any two functions $y_1(t),y_2(t)$.

Since differentiation itself is linear:

\[ L[c_1y_1+c_2y_2]=t^2(c_1y_1''+c_2y_2'')+7t(c_1y_1'+c_2y_2')+8t(c_1y_1+c_2y_2) \]

Regrouping the terms:

\[ =c_1(t^2y_1''+7ty_1'+8ty_1)+c_2(t^2y_2''+7ty_2'+8ty_2)=c_1L[y_1]+c_2L[y_2] \]

The superposition property holds exactly, so $L$ is a linear operator, confirming the equation $L[y]=10\sin t$ is linear.

Now test time invariance by shifting time. If $y(t)$ solves the equation and we define $z(t)=y(t-t_0)$, a genuinely time-invariant system would have $z(t)$ solve the same equation with the input shifted to $10\sin(t-t_0)$. But $y(t-t_0)$ actually satisfies

\[ (t-t_0)^2y''(t-t_0)+7(t-t_0)y'(t-t_0)+8(t-t_0)y(t-t_0)=10\sin(t-t_0) \]

which uses coefficients $(t-t_0)^2,7(t-t_0),8(t-t_0)$, not the original coefficients $t^2,7t,8t$, unless $t_0=0$. Since the shifted function does not satisfy the equation with the original, unshifted coefficients, the system is not time-invariant, only linear.

Combining both checks: the equation is an ordinary (not partial) equation, it is linear, and it is time-varying rather than time-invariant.

\[ \boxed{\text{Linear differential equation}} \]