Consider the differential equation $f^{\prime \prime}+p f^{\prime}+q f=0$ where $p$ and $q$ are cons...

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.

For arbitrary constants $\alpha, \beta$, the differential equation representing the family of curves...

We can verify by differentiating and then eliminating the two constants α, β:Start with$$y=(\alpha x+\beta)e^x.$$Compute its derivatives:$$y'=\bigl(\alpha x+(\alpha+\beta)\bigr)e^x,$$$$y''=\bigl(\alpha x+(2\alpha+\beta)\bigr)e^x.$$Form the combination$$y''-2y'+y =\bigl[\alpha x+(2\alpha+\beta)\bigr]e^x -2\bigl[\alpha x+(\alpha+\beta)\bigr]e^x +(\alpha x+\beta)e^x =0.$$Hence the required differential equation is$$\boxed{y''-2y'+y=0},$$which is Option A.

Which of the following differential equations has $y=e^x$ as one of its particular solutions?...

To determine which differential equation has $ y = e^x $ as a particular solution, we need to verify the expression by substituting $ y = e^x $ and its derivatives into each of the options.First, compute the first and second derivatives of $ y = e^x $:$ \frac{d y}{d x} = e^x, \quad \frac{d^2 y}{d x^2} = e^x $Substitute these derivatives into each differential equation to check if $ y = e^x $ satisfies the expression.Option AFor the equation:$ y \frac{d^2 y}{d x^2} + e^x \frac{d y}{d x} + y^2 = e^{2x} $Substitute the values:$ e^x \cdot e^x + e^x \cdot e^x + (e^x)^2 = e^{2x} $$ e^{2x} + e^{2x} + e^{2x} = 3e^{2x} \neq e^{2x} $Option BFor the equation:$ y \frac{d^2 y}{d x^2} - e^x \frac{d y}{d x} + y^2 = e^{2x} $Substitute the values:$ e^x \cdot e^x - e^x \cdot e^x + (e^x)^2 = e^{2x} $$ e^{2x} - e^{2x} + e^{2x} = e^{2x} $This matches the right side of the equation, so $ y = e^x $ is indeed a particular solution for this equation.Option CFor the equation:$ y \frac{d^2 y}{d x^2} - e^x \frac{d y}{d x} + y^2 = e^x $Substitute the values:$ e^x \cdot e^x - e^x \cdot e^x + (e^x)^2 = e^x $$ e^{2x} - e^{2x} + e^{2x} = e^{2x} \neq e^x $Option DFor the equation:$ y \frac{d^2 y}{d x^2} + e^x \frac{d y}{d x} + y^2 = e^x $Substitute the values:$ e^x \cdot e^x + e^x \cdot e^x + (e^x)^2 = e^x $$ e^{2x} + e^{2x} + e^{2x} = 3e^{2x} \neq e^x $In conclusion, Option B is the differential equation for which $ y = e^x $ is a particular solution.

Consider the differential equation $\cos (y) \frac{d y}{d x}+\frac{1}{x} \sin (y)=x, \quad(x>0)$; gi...

Given the differential equation:$ \cos (y) \frac{d y}{d x} + \frac{1}{x} \sin (y) = x $we'll use the substitution $\sin(y) = t$, which gives us:$ \frac{d t}{d x} = \cos (y) \cdot \frac{d y}{d x} $Substituting in the original equation, we obtain:$ \frac{d t}{d x} + \frac{t}{x} = x $This is a linear first-order differential equation, which can be solved using an integrating factor. The integrating factor is $e^{\int \frac{1}{x} \, dx} = e^{\log(x)} = x$. Multiplying through by the integrating factor gives:$ x \frac{d t}{d x} + t = x^2 $Integrating with respect to $x$ yields:$ t x = \int x^2 \, dx = \frac{x^3}{3} + c $Substitute back for $t$:$ \sin(y) x = \frac{x^3}{3} + c $Given the initial condition $y = \frac{\pi}{2}$ at $x = \sqrt{3}$, we find:$ \sin\left(\frac{\pi}{2}\right) \cdot \sqrt{3} = \frac{(\sqrt{3})^3}{3} + c $This simplifies to:$ \sqrt{3} = \frac{3\sqrt{3}}{3} + c \quad \Rightarrow \quad \sqrt{3} = \sqrt{3} + c \quad \Rightarrow \quad c = 0 $Thus, the final form of the equation is:$ x \sin (y) = \frac{x^3}{3} $or$ y = \sin^{-1}\left(\frac{x^2}{3}\right) $For $x = \sqrt{\frac{3}{2}}$, we find:$ y = \sin^{-1}\left(\frac{\frac{3}{2}}{3}\right) = \sin^{-1}\left(\frac{1}{2}\right) $Thus, the value of $y$ at $x = \sqrt{\frac{3}{2}}$ is:$ y = \frac{\pi}{6} $

Differential Equations - IAT-IISER Maths PYQs

01

PYQ 2020

medium

maths ID: iat-iise

Consider the differential equation where and are constants, and . If y is a nonzero solution of the equation, then which of the following statements is correct?

1

can never be a solution

2

can never be a solution

3

can never be a solution

4

can never be a solution

02

PYQ 2022

easy

maths ID: iat-iise

For arbitrary constants, the differential equation representing the family of curves is

1

2

3

4

03

PYQ 2023

medium

maths ID: iat-iise

Which of the following differential equations has as one of its particular solutions?

1

2

3

4

04

PYQ 2024

medium

maths ID: iat-iise

Consider the differential equation; given that, at . Which one of the following is the value of at ?

1

2

3

4

05

PYQ 2025

medium

maths ID: iat-iise

Which one of the following is the solution of the differential equation$ y=0 x=1$ ?

1

2

3

4

Differential Equations

High-Yield Trend

Questions
5 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Differential Equations

High-Yield Trend

Questions 5 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Questions
5 MCQs