Differential Equations

To determine whether a function is a solution to the differential equation , we need to substitute the function into the differential equation and check if the equation holds true.Steps to Verify:Substitute the function : 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 , to verify if the expression can hold true.Analysis of Options:A) Compute the derivatives: Substitute in the equation: Simplify and group terms: Conclusion: Since is a solution to the original equation, , making a valid solution.B) Compute the derivatives: Substitute in the equation: Simplify and consider possible conditions: This leads to situations where either or , both contradicting the given conditions ( and is non-zero). So, cannot be a solution.Conclusion:Based on this analysis, 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.

We can verify by differentiating and then eliminating the two constants α, β:Start with$ $$ $which is Option A.

To determine which differential equation has as a particular solution, we need to verify the expression by substituting and its derivatives into each of the options.First, compute the first and second derivatives of : Substitute these derivatives into each differential equation to check if satisfies the expression.Option AFor the equation: Substitute the values: Option BFor the equation: Substitute the values: This matches the right side of the equation, so is indeed a particular solution for this equation.Option CFor the equation: Substitute the values: Option DFor the equation: Substitute the values: In conclusion, Option B is the differential equation for which is a particular solution.

Given the differential equation: we'll use the substitution , which gives us: Substituting in the original equation, we obtain: This is a linear first-order differential equation, which can be solved using an integrating factor. The integrating factor is . Multiplying through by the integrating factor gives: Integrating with respect to yields: Substitute back for : Given the initial condition at , we find: This simplifies to: Thus, the final form of the equation is: or For , we find: Thus, the value of at is: