JEE Main 2022 (26 June Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Let the solution curve $y = y (x)$ of the differential equation $(4 + x^{2}) d y - 2 x (x^{2} + 3 y + 4) d x = 0$ pass through the origin. Then $y (2)$ is equal to _______.

Q: Why did we move the $6xy$ term to the left side?

In a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, the dependent variable $y$ must be on the same side as the derivative. By moving the $6xy$ term, we align the equation with the standard form, allowing us to use the powerful Integrating Factor method.

Q: Can't we just use separation of variables here?

Separation of variables works when you can isolate all $x$ terms on one side and all $y$ terms on the other. Here, the $y$ term is trapped inside the expression $2x(x^2 + 3y + 4)$. Because of that additive $3y$, you cannot cleanly separate the variables. The linear form is the only way forward.

Enter Numerical Value:

Visualized Solution

Identify the Differential Equation

Given equation: $(4 + x^{2}) d y - 2 x (x^{2} + 3 y + 4) d x = 0$
Rearranging to find $\frac{d y}{d x}$ :
$(4 + x^{2}) \frac{d y}{d x} = 2 x (x^{2} + 3 y + 4)$

Transform to Standard Linear Form

Expanding the right side: $(x^{2} + 4) \frac{d y}{d x} = 2 x^{3} + 6 x y + 8 x$
Rearranging terms: $(x^{2} + 4) \frac{d y}{d x} - 6 x y = 2 x (x^{2} + 4)$

Isolate $\frac{d y}{d x}$

Dividing by $(x^{2} + 4)$ :
$\frac{d y}{d x} - \frac{6 x}{x ^{2} + 4} y = 2 x$
This is of the form $\frac{d y}{d x} + P (x) y = Q (x)$
Where $P (x) = - \frac{6 x}{x ^{2} + 4}$ and $Q (x) = 2 x$

Calculate the Integrating Factor

I.F. $= e^{\int P (x) d x} = e^{\int - \frac{6 x}{x ^{2} + 4} d x}$
Let $x^{2} + 4 = t \Rightarrow 2 x d x = d t$
$\int - \frac{6 x}{x ^{2} + 4} d x = - 3 \int \frac{d t}{t} = - 3 ln (t)$

Simplify the Integrating Factor

I.F. $= e^{- 3 l n (x^{2} + 4)} = e^{l n (x^{2} + 4)^{- 3}}$
I.F. $= (x^{2} + 4)^{- 3} = \frac{1}{( x ^{2} + 4 ) ^{3}}$

Write the General Solution

General Solution: $y \cdot (I.F.) = \int Q (x) \cdot (I.F.) d x$
$y \cdot \frac{1}{( x ^{2} + 4 ) ^{3}} = \int 2 x \cdot \frac{1}{( x ^{2} + 4 ) ^{3}} d x$

Integrate the Right Hand Side

Let $x^{2} + 4 = t \Rightarrow 2 x d x = d t$
$\int t^{- 3} d t = \frac{t ^{- 2}}{- 2} + C = - \frac{1}{2 t ^{2}} + C$
$\frac{y}{( x ^{2} + 4 ) ^{3}} = - \frac{1}{2 ( x ^{2} + 4 ) ^{2}} + C$

Apply Boundary Condition

Passes through origin $(0, 0)$ :
$0 = - \frac{1}{2 ( 0 ^{2} + 4 ) ^{2}} + C$
$0 = - \frac{1}{2 ( 16 )} + C \Rightarrow C = \frac{1}{32}$

Find the Particular Solution

Substituting $C = \frac{1}{32}$ :
$\frac{y}{( x ^{2} + 4 ) ^{3}} = - \frac{1}{2 ( x ^{2} + 4 ) ^{2}} + \frac{1}{32}$
Multiplying by $(x^{2} + 4)^{3}$ :
$y = - \frac{x ^{2} + 4}{2} + \frac{( x ^{2} + 4 ) ^{3}}{32}$

Calculate $y (2)$

Substitute $x = 2$ :
$y (2) = - \frac{2 ^{2} + 4}{2} + \frac{( 2 ^{2} + 4 ) ^{3}}{32}$
$y (2) = - \frac{8}{2} + \frac{8 ^{3}}{32}$
$y (2) = - 4 + \frac{512}{32} = - 4 + 16$

Final Answer

Final Result:
$y (2) = 12$
The value of $y (2)$ is 12.

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The Sigma Insight: Linear Differential Equations

Solution Diagram

Analyzing the Setup

The given differential equation is:

(4 + x^{2}) d y - 2 x (x^{2} + 3 y + 4) d x = 0

Our first objective is to bring order to chaos by isolating the derivative,

\frac{d y}{d x}

. By rearranging the terms, we obtain:

\frac{d y}{d x} = \frac{2 x ( x ^{2} + 3 y + 4 )}{x ^{2} + 4}

The Art of Rearrangement

Expanding the numerator gives

2 x^{3} + 6 x y + 8 x

. We can group these terms strategically as

2 x (x^{2} + 4) + 6 x y

Dividing this expression by

(x^{2} + 4)

simplifies the equation to:

\frac{d y}{d x} = 2 x + \frac{6 x y}{x ^{2} + 4}

By moving the

y

term to the left, we arrive at the standard linear differential equation form:

\frac{d y}{d x} - (\frac{6 x}{x ^{2} + 4}) y = 2 x

The Magic of the Integrating Factor

We now identify the components of the linear form

\frac{d y}{d x} + P (x) y = Q (x)

, where

P (x) = - \frac{6 x}{x ^{2} + 4}

and

Q (x) = 2 x

. The Integrating Factor (I.F.) is defined as

e^{\int P (x) d x}

We calculate the integral:

\int - \frac{6 x}{x ^{2} + 4} d x = - 3 \int \frac{2 x}{x ^{2} + 4} d x = - 3 ln (x^{2} + 4)

Thus, the I.F. is:

I . F . = e^{- 3 l n (x^{2} + 4)} = (x^{2} + 4)^{- 3} = \frac{1}{( x ^{2} + 4 ) ^{3}}

The Final Integration

The general solution is given by

y \cdot (I . F .) = \int Q (x) \cdot (I . F .) d x

. Substituting our values, we get:

y \cdot \frac{1}{( x ^{2} + 4 ) ^{3}} = \int \frac{2 x}{( x ^{2} + 4 ) ^{3}} d x

Using the substitution

u = x^{2} + 4

, the integral becomes

\int u^{- 3} d u

, which evaluates to:

\frac{u ^{- 2}}{- 2} + C = - \frac{1}{2 ( x ^{2} + 4 ) ^{2}} + C

This yields the general solution:

\frac{y}{( x ^{2} + 4 ) ^{3}} = - \frac{1}{2 ( x ^{2} + 4 ) ^{2}} + C

The Boundary Condition and the Finish Line

The curve passes through the origin

(0, 0)

. Substituting

x = 0

and

y = 0

allows us to solve for

C

0 = - \frac{1}{2 ( 4 ) ^{2}} + C \Rightarrow 0 = - \frac{1}{32} + C \Rightarrow C = \frac{1}{32}

The particular solution is therefore:

y = - \frac{x ^{2} + 4}{2} + \frac{( x ^{2} + 4 ) ^{3}}{32}

To find

y (2)

, we substitute

x = 2

y (2) = - \frac{2 ^{2} + 4}{2} + \frac{( 2 ^{2} + 4 ) ^{3}}{32} = - \frac{8}{2} + \frac{8 ^{3}}{32} = - 4 + \frac{512}{32} = - 4 + 16 = 12

The final answer is 12.

Common Traps & FAQs

Why did we move the

6 x y

term to the left side?

In a linear differential equation of the form

\frac{d y}{d x} + P (x) y = Q (x)

, the dependent variable

y

must be on the same side as the derivative. By moving the

6 x y

term, we align the equation with the standard form, allowing us to use the powerful Integrating Factor method.

Can't we just use separation of variables here?

Separation of variables works when you can isolate all

x

terms on one side and all

y

terms on the other. Here, the

y

term is trapped inside the expression

2 x (x^{2} + 3 y + 4)

. Because of that additive

3 y

, you cannot cleanly separate the variables. The linear form is the only way forward.

How do I know when to use an Integrating Factor?

Whenever you see a differential equation where the dependent variable

y

appears linearly (i.e., just

y

, not

y^{2}

sin (y)

) and its derivative

\frac{d y}{d x}

is present, the Integrating Factor method is your go-to tool. It is the standard approach for solving first-order linear differential equations.

Animated Solution for Mathematics - Differential Equations: Let the solution curve $y = y (x)$ of the differential equation $(4 + x^{2}) d y - 2 x (x^{2} + 3 y + 4) d x = 0$ pass through the origin. Then $y (2)$ is equal to _______.

Enter Numerical Value:

Visualized Solution

Identify the Differential Equation

Transform to Standard Linear Form

Isolate $\frac{d y}{d x}$

Calculate the Integrating Factor

Simplify the Integrating Factor

Write the General Solution

Integrate the Right Hand Side

Apply Boundary Condition

Find the Particular Solution

Calculate $y (2)$

Final Answer

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Art of Rearrangement

The Magic of the Integrating Factor

The Final Integration

The Boundary Condition and the Finish Line

Similar Questions

If the solution of the differential equation $\frac{d y}{d x} + e^{x} (x^{2} - 2) y = (x^{2} - 2 x) (x^{2} - 2) e^{2 x}$ satisfies $y (0) = 0$ , then the value of $y (2)$ is _______.

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

If for $x \geq 0$ , $y = y (x)$ is the solution of the differential equation, $(1 + x) \frac{d y}{d x} = ((1 + x)^{2} + (y - 3))$ , If $y (2) = 0$ then $y (3) = ?$

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 2 y sec^{2} x = 2 sec^{2} x + 3 tan x \cdot sec^{2} x$ such that $y (0) = \frac{5}{4}$ . Then $12 (y (\frac{π}{4}) - e^{- 2})$ is equal to _______.

If for $x \geq 0, y = y (x)$ is the solution of the differential equation $(1 + x) d y = [(1 + x)^{2} + y - 3] d x, y (2) = 0$ , then $y (3)$ is equal to:

If $y = y (x)$ is the solution of the differential equation $4 - x^{2} \frac{d y}{d x} = ((sin^{- 1} (\frac{x}{2}))^{2} - y) sin^{- 1} (\frac{x}{2})$ , $- 2 \leq x \leq 2$ , $y (2) = \frac{π ^{2} - 8}{4}$ , then $y^{2} (0)$ is equal to

If $y = y (x)$ is the solution curve of the differential equation $x^{2} d y + (y - \frac{1}{x}) d x = 0; x > 0$ and $y (1) = 1$ , then $y (\frac{1}{2})$ is equal to :

Let $y = y (x)$ be the solution curve of the differential equation $x (x^{2} + e^{x}) d y + (e^{x} (x - 2) y - x^{3}) d x = 0, x > 0$ passing through the point $(1, 0)$ . Then $y (2)$ is equal to :

A solution of the differential equation $(\frac{d y}{d x})^{2} - x \frac{d y}{d x} + y = 0$ is

Let $y = y (x)$ be the solution of the differential equation $x lo g_{e} x \frac{d y}{d x} + y = x^{2} lo g_{e} x, (x > 1)$ . If $y (2) = 2$ , then $y (e)$ is equal to

Enter Numerical Value:

Visualized Solution

Identify the Differential Equation

Transform to Standard Linear Form

Isolate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​

Calculate the Integrating Factor

Simplify the Integrating Factor

Write the General Solution

Integrate the Right Hand Side

Apply Boundary Condition

Find the Particular Solution

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

Final Answer

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Art of Rearrangement

The Magic of the Integrating Factor

The Final Integration

The Boundary Condition and the Finish Line

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the solution of the differential equation dxdy​+ex(x2−2)y=(x2−2x)(x2−2)e2x satisfies y(0)=0, then the value of y(2) is _______.

.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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If for x≥0, y=y(x) is the solution of the differential equation, (1+x)dxdy​=((1+x)2+(y−3)), If y(2)=0 then y(3)=?

.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​+2ysec2x=2sec2x+3tanx⋅sec2x such that y(0)=45​. Then 12(y(4π​)−e−2) is equal to _______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If for x≥0,y=y(x) is the solution of the differential equation (1+x)dy=[(1+x)2+y−3]dx,y(2)=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 y=y(x) is the solution of the differential equation 4−x2​dxdy​=((sin−1(2x​))2−y)sin−1(2x​) , −2≤x≤2, y(2)=4π2−8​ , then y2(0) is equal to

.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 curve of the differential equation x2dy+(y−x1​)dx=0;x>0 and y(1)=1, then y(21​) 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 x(x2+ex)dy+(ex(x−2)y−x3)dx=0,x>0 passing through the point (1,0). Then y(2) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A solution of the differential equation (dxdy​)2−xdxdy​+y=0 is

.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 xloge​xdxdy​+y=x2loge​x,(x>1). If y(2)=2, then y(e) is equal to

Isolate $\frac{d y}{d x}$

Calculate $y (2)$

If the solution of the differential equation $\frac{d y}{d x} + e^{x} (x^{2} - 2) y = (x^{2} - 2 x) (x^{2} - 2) e^{2 x}$ satisfies $y (0) = 0$ , then the value of $y (2)$ is _______.

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

If for $x \geq 0$ , $y = y (x)$ is the solution of the differential equation, $(1 + x) \frac{d y}{d x} = ((1 + x)^{2} + (y - 3))$ , If $y (2) = 0$ then $y (3) = ?$

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 2 y sec^{2} x = 2 sec^{2} x + 3 tan x \cdot sec^{2} x$ such that $y (0) = \frac{5}{4}$ . Then $12 (y (\frac{π}{4}) - e^{- 2})$ is equal to _______.

If for $x \geq 0, y = y (x)$ is the solution of the differential equation $(1 + x) d y = [(1 + x)^{2} + y - 3] d x, y (2) = 0$ , then $y (3)$ is equal to:

If $y = y (x)$ is the solution of the differential equation $4 - x^{2} \frac{d y}{d x} = ((sin^{- 1} (\frac{x}{2}))^{2} - y) sin^{- 1} (\frac{x}{2})$ , $- 2 \leq x \leq 2$ , $y (2) = \frac{π ^{2} - 8}{4}$ , then $y^{2} (0)$ is equal to

If $y = y (x)$ is the solution curve of the differential equation $x^{2} d y + (y - \frac{1}{x}) d x = 0; x > 0$ and $y (1) = 1$ , then $y (\frac{1}{2})$ is equal to :

Let $y = y (x)$ be the solution curve of the differential equation $x (x^{2} + e^{x}) d y + (e^{x} (x - 2) y - x^{3}) d x = 0, x > 0$ passing through the point $(1, 0)$ . Then $y (2)$ is equal to :

A solution of the differential equation $(\frac{d y}{d x})^{2} - x \frac{d y}{d x} + y = 0$ is

Let $y = y (x)$ be the solution of the differential equation $x lo g_{e} x \frac{d y}{d x} + y = x^{2} lo g_{e} x, (x > 1)$ . If $y (2) = 2$ , then $y (e)$ is equal to