Let $y = y(x)$ be the solution of the differential equation $xdy = (y + x^3 \cos x) dx$ with $y(\pi) = 0$, then $y\left(\frac{\pi}{2}\right)$ is equal to: Options: (A) $\frac{\pi^2}{4} + \frac{\pi}{2}$ (B) $\frac{\pi^2}{2} + \frac{\pi}{4}$ (C) $\frac{\pi^2}{2} - \frac{\pi}{4}$ (D) $\frac{\pi^2}{4} - \frac{\pi}{2}$

### Analyzing the Setup We are given the differential equation $x dy = (y + x^3 \cos x) dx$ with the initial condition $y(\pi) = 0$. Our goal is to determine the value of $y(\pi/2)$. First, we rearrange the terms to isolate the differential components: $$x dy - y dx = x^3 \cos x dx$$ ### The Master Equation The expression on the left side, $x dy - y dx$, is the numerator of the derivative of the quotient $\frac{y}{x}$. To utilize this, we divide the entire equation by $x^2$: $$\frac{x dy - y dx}{x^2} = \frac{x^3 \cos x dx}{x^2}$$ This simplifies elegantly to the differential form: $$d\left(\frac{y}{x}\right) = x \cos x dx$$ ### Integration and General Solution We now integrate both sides of the equation. The left side integrates directly to $\frac{y}{x}$, while the right side requires integration by parts: $$\int d\left(\frac{y}{x}\right) = \int x \cos x dx$$ Using the ILATE rule, we set $u = x$ and $dv = \cos x dx$. This yields: $$\frac{y}{x} = x \sin x - \int \sin x dx = x \sin x + \cos x + C$$ Multiplying by $x$, we obtain the general solution: $$y = x^2 \sin x + x \cos x + Cx$$ ### Applying Initial Conditions We apply the condition $y(\pi) = 0$ to solve for the constant $C$. Substituting $x = \pi$ and $y = 0$: $$0 = \pi^2 \sin(\pi) + \pi \cos(\pi) + C\pi$$ Since $\sin(\pi) = 0$ and $\cos(\pi) = -1$, the equation becomes: $$0 = 0 - \pi + C\pi \Rightarrow C = 1$$ ### Final Calculation Substituting $C = 1$ back into our general solution, we get: $$y = x^2 \sin x + x \cos x + x$$ To find $y(\pi/2)$, we evaluate the expression at $x = \pi/2$: $$y\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^2 \sin\left(\frac{\pi}{2}\right) + \left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) + \frac{\pi}{2}$$ Given $\sin(\pi/2) = 1$ and $\cos(\pi/2) = 0$, the final result is: **$$y\left(\frac{\pi}{2}\right) = \frac{\pi^2}{4} + \frac{\pi}{2}$$**

JEE Main 2021 (25 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $x d y = (y + x^{3} cos x) d x$ with $y (π) = 0$ , then $y (\frac{π}{2})$ is equal to:

Q: Why did we divide by $x^2$ specifically? Could we have divided by something else?

Great question! We chose $x^2$ because it is the denominator required by the quotient rule for the function $\frac{y}{x}$. If we had divided by $x$, we would have had $\frac{x dy - y dx}{x}$, which doesn't simplify into a clean derivative. Always look for the denominator that completes the differential form you are aiming for.

Q: Can I solve this using the integrating factor method for linear differential equations instead?

Absolutely! You can rewrite the equation as $\frac{dy}{dx} - \frac{1}{x}y = x^2 \cos x$. The integrating factor would be $e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x}$. Multiplying by this factor leads to the same result. The exact differential method is just a faster, more elegant shortcut!

Q: Why does the constant $C$ appear, and how do I know it's correct?

The constant $C$ is the signature of an indefinite integral. Since we are solving a first-order differential equation, we expect one arbitrary constant. We determine its specific value using the initial condition $y(\pi) = 0$. Think of the general solution as a family of curves, and the initial condition as the specific pin that selects the one curve passing through your required point.

Select Answer:

Visualized Solution

Given Differential Equation

Given differential equation: $x d y = (y + x^{3} cos x) d x$
Initial condition: $y (π) = 0$
Target: Find $y (\frac{π}{2})$

Rearranging the Terms

Expand the right side: $x d y = y d x + x^{3} cos x d x$
Rearrange: $x d y - y d x = x^{3} cos x d x$

Dividing by $x^{2}$

Divide the entire equation by $x^{2}$ :
$\frac{x d y - y d x}{x ^{2}} = \frac{x ^{3} c o s x d x}{x ^{2}}$
Simplify the right side: $\frac{x d y - y d x}{x ^{2}} = x cos x d x$

Recognizing the Exact Differential

Recall the quotient rule: $d (\frac{u}{v}) = \frac{v d u - u d v}{v ^{2}}$
Substitute $u = y$ and $v = x$ : $d (\frac{y}{x}) = \frac{x d y - y d x}{x ^{2}}$
The equation becomes: $d (\frac{y}{x}) = x cos x d x$

Integrating Both Sides

Integrate both sides: $\int d (\frac{y}{x}) = \int x cos x d x$
Left side: $\frac{y}{x}$
Right side: Apply Integration by Parts $\int uv d x = u \int v d x - \int (u^{'} \int v d x) d x$

Applying Integration by Parts

Let $u = x$ and $d v = cos x d x$
Then $d u = d x$ and $v = sin x$
$\int x cos x d x = x sin x - \int sin x d x$
$\int x cos x d x = x sin x - (- cos x) + C$
$\int x cos x d x = x sin x + cos x + C$

General Solution for $\frac{y}{x}$

General solution: $\frac{y}{x} = x sin x + cos x + C$
Multiply by $x$ : $y = x^{2} sin x + x cos x + C x$

Applying Initial Condition $y (π) = 0$

Substitute $x = π$ and $y = 0$ into $y = x^{2} sin x + x cos x + C x$ :
$0 = π^{2} sin (π) + π cos (π) + C π$
Since $sin (π) = 0$ and $cos (π) = - 1$ :
$0 = π^{2} (0) + π (- 1) + C π$

Solving for Constant $C$

$0 = - π + C π$
$π = C π$
$C = 1$

Finding the Specific Solution $y (x)$

Substitute $C = 1$ back into the general solution:
$y = x^{2} sin x + x cos x + (1) x$
Specific solution: $y = x^{2} sin x + x cos x + x$

Calculating $y (\frac{π}{2})$

Substitute $x = \frac{π}{2}$ :
$y (\frac{π}{2}) = (\frac{π}{2})^{2} sin (\frac{π}{2}) + \frac{π}{2} cos (\frac{π}{2}) + \frac{π}{2}$
Use $sin (\frac{π}{2}) = 1$ and $cos (\frac{π}{2}) = 0$ :
$y (\frac{π}{2}) = \frac{π ^{2}}{4} (1) + \frac{π}{2} (0) + \frac{π}{2}$

Final Answer and Summary

Final result: $y (\frac{π}{2}) = \frac{π ^{2}}{4} + \frac{π}{2}$
Key Takeaways:
Recognizing exact differentials like $d (\frac{y}{x})$ can simplify DEs significantly.
Always use initial conditions to find the specific constant $C$ .
Integration by Parts is a frequent tool in JEE calculus problems.

00:00 / 00:00

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

We are given the differential equation

x d y = (y + x^{3} cos x) d x

with the initial condition

y (π) = 0

. Our goal is to determine the value of

y (π /2)

First, we rearrange the terms to isolate the differential components:

x d y - y d x = x^{3} cos x d x

The Master Equation

The expression on the left side,

x d y - y d x

, is the numerator of the derivative of the quotient

\frac{y}{x}

. To utilize this, we divide the entire equation by

x^{2}

\frac{x d y - y d x}{x ^{2}} = \frac{x ^{3} cos x d x}{x ^{2}}

This simplifies elegantly to the differential form:

d (\frac{y}{x}) = x cos x d x

Integration and General Solution

We now integrate both sides of the equation. The left side integrates directly to

\frac{y}{x}

, while the right side requires integration by parts:

\int d (\frac{y}{x}) = \int x cos x d x

Using the ILATE rule, we set

u = x

and

d v = cos x d x

. This yields:

\frac{y}{x} = x sin x - \int sin x d x = x sin x + cos x + C

Multiplying by

x

, we obtain the general solution:

y = x^{2} sin x + x cos x + C x

Applying Initial Conditions

We apply the condition

y (π) = 0

to solve for the constant

C

. Substituting

x = π

and

y = 0

0 = π^{2} sin (π) + π cos (π) + C π

Since

sin (π) = 0

and

cos (π) = - 1

, the equation becomes:

0 = 0 - π + C π \Rightarrow C = 1

Final Calculation

Substituting

C = 1

back into our general solution, we get:

y = x^{2} sin x + x cos x + x

To find

y (π /2)

, we evaluate the expression at

x = π /2

y (\frac{π}{2}) = (\frac{π}{2})^{2} sin (\frac{π}{2}) + (\frac{π}{2}) cos (\frac{π}{2}) + \frac{π}{2}

Given

sin (π /2) = 1

and

cos (π /2) = 0

, the final result is: $y (\frac{π}{2}) = \frac{π ^{2}}{4} + \frac{π}{2}$

Common Traps & FAQs

Why did we divide by

x^{2}

specifically? Could we have divided by something else?

Great question! We chose

x^{2}

because it is the denominator required by the quotient rule for the function

\frac{y}{x}

. If we had divided by

x

, we would have had

\frac{x d y - y d x}{x}

, which doesn't simplify into a clean derivative. Always look for the denominator that completes the differential form you are aiming for.

Can I solve this using the integrating factor method for linear differential equations instead?

Absolutely! You can rewrite the equation as

\frac{d y}{d x} - \frac{1}{x} y = x^{2} cos x

. The integrating factor would be

e^{\int - \frac{1}{x} d x} = e^{- l n x} = \frac{1}{x}

. Multiplying by this factor leads to the same result. The exact differential method is just a faster, more elegant shortcut!

Why does the constant

C

appear, and how do I know it's correct?

The constant

C

is the signature of an indefinite integral. Since we are solving a first-order differential equation, we expect one arbitrary constant. We determine its specific value using the initial condition

y (π) = 0

. Think of the general solution as a family of curves, and the initial condition as the specific pin that selects the one curve passing through your required point.

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $x d y = (y + x^{3} cos x) d x$ with $y (π) = 0$ , then $y (\frac{π}{2})$ is equal to:

Select Answer:

Visualized Solution

Given Differential Equation

Rearranging the Terms

Dividing by $x^{2}$

Recognizing the Exact Differential

Integrating Both Sides

Applying Integration by Parts

General Solution for $\frac{y}{x}$

Applying Initial Condition $y (π) = 0$

Solving for Constant $C$

Finding the Specific Solution $y (x)$

Calculating $y (\frac{π}{2})$

Final Answer and Summary

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Master Equation

Integration and General Solution

Applying Initial Conditions

Final Calculation

Similar Questions

Let $y = y (x)$ be the solution of the differential equation $cos x (3 sin x + cos x + 3) d y = (1 + y sin x (3 sin x + cos x + 3)) d x$ , $0 \leq x \leq \frac{π}{2}, y (0) = 0$ . Then $y (\frac{π}{3})$ is equal to:

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 3 (tan^{2} x) y + 3 y = sec^{2} x$ , $y (0) = \frac{1}{3} + e^{3}$ . Then $y (\frac{π}{4})$ is equal to

Let $y = y (x)$ be the solution of the differential equation $2 cos x \frac{d y}{d x} = sin 2 x - 4 y sin x, x \in (0, \frac{π}{2}) .$ If $y (\frac{π}{3}) = 0$ then $y^{'} (\frac{π}{4}) + y (\frac{π}{4})$ is equal to

Let $y = y (x)$ be the solution of the differential equation $x^{4} d y + (4 x^{3} y + 2 sin x) d x = 0, x > 0, y (\frac{π}{2}) = 0$ . Then $π^{4} y (\frac{π}{3})$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $cos x \frac{d y}{d x} + 2 y sin x = sin 2 x$ , $x \in (0, \frac{π}{2})$ . If $y (π /3) = 0$ , then $y (π /4)$ is equal to :

Let $y = y (x)$ be the solution of the differential equation, $x y^{'} - y = x^{2} (x cos x + sin x)$ , $x > 0$ . If $y (π) = π$ , then $y^{''} (\frac{π}{2}) + y (\frac{π}{2})$ is equal to :

Let $y = y (x)$ be the solution curve of the differential equation $sec y \frac{d y}{d x} + 2 x sin y = x^{3} cos y, y (1) = 0$ . Then $y (3)$ is equal to :

If $\frac{d y}{d x} + \frac{3}{c o s ^{2} x} y = \frac{1}{c o s ^{2} x}, x \in (- \frac{π}{3}, \frac{π}{3})$ , and $y (\frac{π}{4}) = \frac{4}{3}$ , then $y (- \frac{π}{4})$ equals :

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + (tan x) y = sin x, 0 \leq x \leq \frac{π}{3}$ , with $y (0) = 0$ , then $y (\frac{π}{4})$ equal to :

Let $y = y (x)$ be the solution of the differential equation $(x^{2} + 4)^{2} d y + (2 x^{3} y + 8 x y - 2) d x = 0$ . If $y (0) = 0$ , then $y (2)$ is equal to

Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation xdy=(y+x3cosx)dx with y(π)=0, then y(2π​) is equal to:

Select Answer:

Visualized Solution

Given Differential Equation

Rearranging the Terms

Dividing by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2

Recognizing the Exact Differential

Integrating Both Sides

Applying Integration by Parts

General Solution for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xy​

Applying Initial Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(π)=0

Solving for Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Finding the Specific Solution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(x)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(2π​)

Final Answer and Summary

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Master Equation

Integration and General Solution

Applying Initial Conditions

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation cosx(3sinx+cosx+3)dy=(1+ysinx(3sinx+cosx+3))dx, 0≤x≤2π​,y(0)=0. Then y(3π​) is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation dxdy​+3(tan2x)y+3y=sec2x, y(0)=31​+e3. Then y(4π​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation 2 cos xdxdy​=sin 2x−4y sin x,x∈(0,2π​). If y(3π​)=0 then y′(4π​)+y(4π​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation x4dy+(4x3y+2sinx)dx=0,x>0,y(2π​)=0. Then π4y(3π​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation cosxdxdy​+2ysinx=sin2x, x∈(0,2π​). If y(π/3)=0, then y(π/4) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation, xy′−y=x2(xcosx+sinx), x>0. If y(π)=π, then y′′(2π​)+y(2π​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution curve of the differential equation secydxdy​+2xsiny=x3cosy,y(1)=0. Then y(3​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If dxdy​+cos2x3​y=cos2x1​,x∈(−3π​,3π​), and y(4π​)=34​, then y(−4π​) equals :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x) is the solution of the differential equation dxdy​+(tanx)y=sinx,0≤x≤3π​, with y(0)=0, then y(4π​) equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation (x2+4)2dy+(2x3y+8xy−2)dx=0. If y(0)=0, then y(2) is equal to

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ be the solution of the differential equation $x d y = (y + x^{3} cos x) d x$ with $y (π) = 0$ , then $y (\frac{π}{2})$ is equal to:

Dividing by $x^{2}$

General Solution for $\frac{y}{x}$

Applying Initial Condition $y (π) = 0$

Solving for Constant $C$

Finding the Specific Solution $y (x)$

Calculating $y (\frac{π}{2})$

Let $y = y (x)$ be the solution of the differential equation $cos x (3 sin x + cos x + 3) d y = (1 + y sin x (3 sin x + cos x + 3)) d x$ , $0 \leq x \leq \frac{π}{2}, y (0) = 0$ . Then $y (\frac{π}{3})$ is equal to:

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 3 (tan^{2} x) y + 3 y = sec^{2} x$ , $y (0) = \frac{1}{3} + e^{3}$ . Then $y (\frac{π}{4})$ is equal to

Let $y = y (x)$ be the solution of the differential equation $2 cos x \frac{d y}{d x} = sin 2 x - 4 y sin x, x \in (0, \frac{π}{2}) .$ If $y (\frac{π}{3}) = 0$ then $y^{'} (\frac{π}{4}) + y (\frac{π}{4})$ is equal to

Let $y = y (x)$ be the solution of the differential equation $x^{4} d y + (4 x^{3} y + 2 sin x) d x = 0, x > 0, y (\frac{π}{2}) = 0$ . Then $π^{4} y (\frac{π}{3})$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $cos x \frac{d y}{d x} + 2 y sin x = sin 2 x$ , $x \in (0, \frac{π}{2})$ . If $y (π /3) = 0$ , then $y (π /4)$ is equal to :

Let $y = y (x)$ be the solution of the differential equation, $x y^{'} - y = x^{2} (x cos x + sin x)$ , $x > 0$ . If $y (π) = π$ , then $y^{''} (\frac{π}{2}) + y (\frac{π}{2})$ is equal to :

Let $y = y (x)$ be the solution curve of the differential equation $sec y \frac{d y}{d x} + 2 x sin y = x^{3} cos y, y (1) = 0$ . Then $y (3)$ is equal to :

If $\frac{d y}{d x} + \frac{3}{c o s ^{2} x} y = \frac{1}{c o s ^{2} x}, x \in (- \frac{π}{3}, \frac{π}{3})$ , and $y (\frac{π}{4}) = \frac{4}{3}$ , then $y (- \frac{π}{4})$ equals :

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + (tan x) y = sin x, 0 \leq x \leq \frac{π}{3}$ , with $y (0) = 0$ , then $y (\frac{π}{4})$ equal to :

Let $y = y (x)$ be the solution of the differential equation $(x^{2} + 4)^{2} d y + (2 x^{3} y + 8 x y - 2) d x = 0$ . If $y (0) = 0$ , then $y (2)$ is equal to