To determine whether a function is a solution to the differential equation $f'' + p f' + q f = 0$, we need to substitute the function into the differential equation and check if the equation holds true.

Steps to Verify:

Substitute the function $f$: Let's start by substituting the candidate function into the differential equation.

Calculate Derivatives: Compute the first and second derivatives of the function.

Substitute Derivatives into the Equation: Substitute the calculated derivatives back into the differential equation.

Simplify: Simplify the resulting expression and check if it equates to zero.

Verify Conditions: Use given conditions, such as $p^2 = 4q$, to verify if the expression can hold true.

Analysis of Options:

A) $f = (1+x)y$

Compute the derivatives:

$ f' = y + (1+x)y' $

$ f'' = (1+x)y'' + 2y' $

Substitute in the equation:

$ (1+x)y'' + 2y' + p[y + (1+x)y'] + q(1+x)y = 0 $

Simplify and group terms:

$ (1+x)\left(y'' + py' + qy\right) + 2y' + py = 0 $

Conclusion: Since $y$ is a solution to the original equation, $y'' + py' + qy = 0$, making $(1+x)y$ a valid solution.

B) $f = y^2$

Compute the derivatives:

$ f' = 2yy' $

$ f'' = 2(y')^2 + 2yy'' $

Substitute in the equation:

$ 2(y')^2 + 2yy'' + p(2yy') + qy^2 = 0 $

Simplify and consider possible conditions:

$ 2(y')^2 + 2\left(y'' + py' + qy\right)y - qy^2 = 0 $

This leads to situations where either $q = 0$ or $y = 0$, both contradicting the given conditions ($q \neq 0$ and $y$ is non-zero). So, $y^2$ cannot be a solution.

Conclusion:

Based on this analysis, $y^2$ can never be a solution to the differential equation given the constraints and defined conditions. You can attempt the remaining options using similar procedures to verify their validity.

Differential Equations

Continue this paper

Similar Differential Equations PYQs

More practice from this question