How did you know that $(x+2)\phi'(x) + \phi(x)$ is the derivative of $(x+2)\phi(x)$?

This is a classic pattern recognition skill in calculus! If you look at the product rule, the derivative of $u(x)v(x)$ is $u'(x)v(x) + u(x)v'(x)$. Here, if we let $u(x) = (x+2)$ and $v(x) = \phi(x)$, then $u'(x) = 1$. So, the derivative is $1 \cdot \phi(x) + (x+2) \cdot \phi'(x)$, which matches our expression perfectly. Always keep an eye out for these hidden product rule structures!

JEE Main 2021 (31 Aug Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: If $x ϕ (x) = \int_{5}^{x} (3 t^{2} - 2 ϕ^{'} (t)) d t, x > - 2$ , and $ϕ (0) = 4$ then $ϕ (2)$ is.

Q: What happens to the constant of integration $C$ after we integrate?

The constant $C$ is essential! When we integrate both sides, we get an indefinite integral, which always introduces a constant. We use the initial condition $\phi(0) = 4$ provided in the problem to solve for $C$. By substituting $x=0$ into our equation $(x+2)\phi(x) = x^3 + C$, we can easily find that $C=8$.

Enter Numerical Value:

Progressive Hints

Hint 1

The function

ϕ (x)

is trapped inside an integral with a variable upper limit

x

. To isolate

ϕ (x)

, you must eliminate the integral. Recall that the fundamental theorem of calculus, specifically the Newton-Leibniz rule, allows you to differentiate an integral with respect to its upper limit.

Hint 2

Once you differentiate both sides, you will encounter terms involving

ϕ^{'} (x)

. On the left-hand side, apply the product rule to

\frac{d}{d x} (x ϕ (x))

. On the right-hand side, apply the Newton-Leibniz rule to

\int_{5}^{x} (3 t^{2} - 2 ϕ^{'} (t)) d t

. This will transform the integral equation into a differential equation.

Hint 3

After differentiating and rearranging, you will arrive at the equation

(x + 2) ϕ^{'} (x) + ϕ (x) = 3 x^{2}

. Observe that the left-hand side is the result of the product rule applied to a specific function. Recognize this as the exact derivative of

(x + 2) ϕ (x)

and integrate both sides to solve for

ϕ (x)

Visualized Solution

Analyze the Given Integral Equation

Given equation: $x ϕ (x) = \int_{5}^{x} (3 t^{2} - 2 ϕ^{'} (t)) d t$
Constraint: $x > - 2$
Initial condition: $ϕ (0) = 4$
Goal: Find the value of $ϕ (2)$

The Tool: Newton-Leibniz Rule

To remove the integral, we differentiate both sides with respect to $x$ .
Newton-Leibniz Rule: $\frac{d}{d x} \int_{a}^{g (x)} f (t) d t = f (g (x)) \cdot g^{'} (x)$
Here, upper limit is $x$ , so its derivative is $1$ .

Differentiating the Left-Hand Side

Applying Product Rule to the LHS: $\frac{d}{d x} (x ϕ (x))$
$\frac{d}{d x} (x ϕ (x)) = x \cdot \frac{d}{d x} (ϕ (x)) + ϕ (x) \cdot \frac{d}{d x} (x)$
Result: $ϕ (x) + x ϕ^{'} (x)$

Differentiating the Right-Hand Side

Applying Leibniz Rule to the RHS: $\frac{d}{d x} \int_{5}^{x} (3 t^{2} - 2 ϕ^{'} (t)) d t$
Substitute $t = x$ into the integrand.
Result: $3 x^{2} - 2 ϕ^{'} (x)$

Equating and Rearranging

Equating LHS and RHS derivatives: $ϕ (x) + x ϕ^{'} (x) = 3 x^{2} - 2 ϕ^{'} (x)$
Group $ϕ^{'} (x)$ terms together: $x ϕ^{'} (x) + 2 ϕ^{'} (x) + ϕ (x) = 3 x^{2}$
Factor out $ϕ^{'} (x)$ : $(x + 2) ϕ^{'} (x) + ϕ (x) = 3 x^{2}$

Recognizing the Exact Differential

Notice the LHS: $(x + 2) ϕ^{'} (x) + ϕ (x)$
This is the exact derivative of a product: $\frac{d}{d x} ((x + 2) ϕ (x))$
Because $\frac{d}{d x} (x + 2) = 1$ , the expression matches $(x + 2) ϕ^{'} (x) + ϕ (x) \cdot 1$
Simplified equation: $\frac{d}{d x} ((x + 2) ϕ (x)) = 3 x^{2}$

Integrating Both Sides

Integrate both sides with respect to $x$ : $\int \frac{d}{d x} ((x + 2) ϕ (x)) d x = \int 3 x^{2} d x$
The integral of a derivative gives back the original function.
$\int 3 x^{2} d x = x^{3} + C$
Result: $(x + 2) ϕ (x) = x^{3} + C$

Applying the Initial Condition

We need to find the constant $C$ .
Use the given initial condition: $ϕ (0) = 4$
Substitute $x = 0$ into the equation: $(0 + 2) ϕ (0) = 0^{3} + C$

Finding the Constant $C$

Substitute $ϕ (0) = 4$ : $2 \cdot 4 = C$
$8 = C$
The equation becomes: $(x + 2) ϕ (x) = x^{3} + 8$

Finding the Expression for $ϕ (x)$

Solve for $ϕ (x)$ by dividing both sides by $(x + 2)$ .
$ϕ (x) = \frac{x ^{3} + 8}{x + 2}$
Note: The problem states $x > - 2$ , so $x + 2 \neq = 0$ , making division valid.

Simplifying the Function

Recognize the sum of cubes in the numerator: $x^{3} + 8 = x^{3} + 2^{3}$
Use the identity $a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$
$x^{3} + 8 = (x + 2) (x^{2} - 2 x + 4)$
Substitute back: $ϕ (x) = \frac{( x + 2 ) ( x ^{2} - 2 x + 4 )}{x + 2}$
Cancel $(x + 2)$ : $ϕ (x) = x^{2} - 2 x + 4$

Calculating the Final Answer

We need to find $ϕ (2)$ .
Substitute $x = 2$ into the simplified function: $ϕ (2) = 2^{2} - 2 (2) + 4$
$ϕ (2) = 4 - 4 + 4$
$ϕ (2) = 4$
Final Answer is $4$ .

00:00 / 00:00

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The problem presents us with the integral equation:

x ϕ (x) = \int_{5}^{x} (3 t^{2} - 2 ϕ^{'} (t)) d t

The function

ϕ (x)

is trapped within an integral, accompanied by its derivative

ϕ^{'} (t)

. To solve this, we must first liberate the function using the Fundamental Theorem of Calculus.

The Liberation

We differentiate both sides of the equation with respect to

x

. Applying the product rule to the left-hand side, we obtain:

\frac{d}{d x} (x ϕ (x)) = ϕ (x) + x ϕ^{'} (x)

Applying the Newton-Leibniz rule to the right-hand side, the integral vanishes, leaving the integrand evaluated at

x

ϕ (x) + x ϕ^{'} (x) = 3 x^{2} - 2 ϕ^{'} (x)

The Algebraic Dance

Next, we group the terms involving

ϕ^{'} (x)

to simplify the expression. Moving

- 2 ϕ^{'} (x)

to the left-hand side yields:

x ϕ^{'} (x) + 2 ϕ^{'} (x) + ϕ (x) = 3 x^{2}

Factoring out

ϕ^{'} (x)

, we arrive at the following compact form:

(x + 2) ϕ^{'} (x) + ϕ (x) = 3 x^{2}

The Hidden Symmetry

Observe that the left-hand side is the result of the product rule applied to the function

(x + 2) ϕ (x)

. Specifically, since the derivative of

(x + 2)

1

, we can rewrite the equation as:

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

This transformation simplifies the differential equation into a direct integration problem.

The Final Reveal

Integrating both sides with respect to

x

, we obtain:

(x + 2) ϕ (x) = \int 3 x^{2} d x = x^{3} + C

To determine the constant

C

, we evaluate the original integral equation at

x = 5

. This gives

5 ϕ (5) = \int_{5}^{5} (...) d t = 0

, implying

ϕ (5) = 0

. Substituting

x = 5

into our integrated equation:

(5 + 2) ϕ (5) = 5^{3} + C \Rightarrow 7 (0) = 125 + C \Rightarrow C = - 125

Thus, the function is defined by:

(x + 2) ϕ (x) = x^{3} - 125

Using the difference of cubes identity,

x^{3} - 125 = (x - 5) (x^{2} + 5 x + 25)

, we see that

ϕ (x) = \frac{x ^{3} - 125}{x + 2}

. Finally, we calculate the requested value:

ϕ (2) = \frac{2 ^{3} - 125}{2 + 2} = \frac{8 - 125}{4} = \frac{- 117}{4}

The final value is $- 29.25$ .

Common Traps & FAQs

Why do we differentiate the equation? Can't we just integrate the right-hand side directly?

That is a great question! The problem is that the right-hand side contains

ϕ^{'} (t)

inside the integral. Since we do not know the function

ϕ (t)

yet, we cannot integrate it directly. By differentiating both sides with respect to

x

, we convert the integral equation into a differential equation, which is much easier to solve.

How did you know that

(x + 2) ϕ^{'} (x) + ϕ (x)

is the derivative of

(x + 2) ϕ (x)

This is a classic pattern recognition skill in calculus! If you look at the product rule, the derivative of

u (x) v (x)

u^{'} (x) v (x) + u (x) v^{'} (x)

. Here, if we let

u (x) = (x + 2)

and

v (x) = ϕ (x)

, then

u^{'} (x) = 1

. So, the derivative is

1 \cdot ϕ (x) + (x + 2) \cdot ϕ^{'} (x)

, which matches our expression perfectly. Always keep an eye out for these hidden product rule structures!

What happens to the constant of integration

C

after we integrate?

The constant

C

is essential! When we integrate both sides, we get an indefinite integral, which always introduces a constant. We use the initial condition

ϕ (0) = 4

provided in the problem to solve for

C

. By substituting

x = 0

into our equation

(x + 2) ϕ (x) = x^{3} + C

, we can easily find that

C = 8

Animated Solution for Mathematics - Differential Equations: If $x ϕ (x) = \int_{5}^{x} (3 t^{2} - 2 ϕ^{'} (t)) d t, x > - 2$ , and $ϕ (0) = 4$ then $ϕ (2)$ is.

Enter Numerical Value:

Visualized Solution

Analyze the Given Integral Equation

The Tool: Newton-Leibniz Rule

Differentiating the Left-Hand Side

Differentiating the Right-Hand Side

Equating and Rearranging

Recognizing the Exact Differential

Integrating Both Sides

Applying the Initial Condition

Finding the Constant $C$

Finding the Expression for $ϕ (x)$

Simplifying the Function

Calculating the Final Answer

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Liberation

The Algebraic Dance

The Hidden Symmetry

The Final Reveal

Similar Questions

Let $F : [3, 5] \to R$ be a twice differentiable function on $(3, 5)$ such that $F (x) = e^{- x} \int_{3}^{x} (3 t^{2} + 2 t + 4 F^{'} (t)) d t$ . If $F^{'} (4) = \frac{α e ^{β} - 224}{( e ^{β} - 4 ) ^{2}}$ , then $α + β$ is equal to

Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function such that $f (1) = 2$ . If $6 \int_{1}^{x} f (t) d t = 3 x f (x) - x^{3}$ for all $x \geq 1$ , then the value of $f (2)$ is .........

Let $y = f (x)$ be the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{6} + 4 x}{1 - x ^{2}},$ $- 1 < x < 1$ such that $f (0) = 0$ . If $6 \int_{- 1/2}^{1/2} f (x) d x = 2 π - α$ then $α^{2}$ is equal to

Let f be a differentiable function such that $2 (x + 2)^{2} f (x) - 3 (x + 2)^{2} = 10 \int_{0}^{x} (t + 2) f (t) d t, x \geq 0$ Then $f (2)$ 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

Let $f : R \to R$ be a differentiable function such that $f^{'} (x) + f (x) = \int_{0}^{2} f (t) d t$ . If $f (0) = e^{- 2}$ , then $2 f (0) - f (2)$ is equal to ______.

If $\frac{d x}{d y} = \frac{1 + x - y ^{2}}{y}, x (1) = 1$ , then $5 x (2)$ is equal to :

If $x = x (t)$ is the solution of the differential equation $(t + 1) d x = (2 x + (t + 1)^{4}) d t, x (0) = 2$ , then, $x (1)$ equals

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

Let $f : [1, \infty) \to R$ be a differentiable function such that $f (1) = \frac{1}{3}$ and $3 \int_{1}^{x} f (t) d t = x f (x) - \frac{x ^{3}}{3}, x \in [1, \infty)$ . Let $e$ denote the base of the natural logarithm. Then the value of $f (e)$ is

Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If xϕ(x)=∫5x​(3t2−2ϕ′(t))dt,x>−2, and ϕ(0)=4 then ϕ(2) is.

Enter Numerical Value:

Visualized Solution

Analyze the Given Integral Equation

The Tool: Newton-Leibniz Rule

Differentiating the Left-Hand Side

Differentiating the Right-Hand Side

Equating and Rearranging

Recognizing the Exact Differential

Integrating Both Sides

Applying the Initial Condition

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

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

Simplifying the Function

Calculating the Final Answer

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Liberation

The Algebraic Dance

The Hidden Symmetry

The Final Reveal

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let F:[3,5]→R be a twice differentiable function on (3,5) such that F(x)=e−x∫3x​(3t2+2t+4F′(t))dt. If F′(4)=(eβ−4)2αeβ−224​, then α+β is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:[1,∞)→[2,∞) be a differentiable function such that f(1)=2. If 6∫1x​f(t)dt=3xf(x)−x3 for all x≥1, then the value of f(2) is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f be a differentiable function such that 2(x+2)2f(x)−3(x+2)2=10∫0x​(t+2)f(t)dt,x≥0 Then f(2) is equal to

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a differentiable function such that f′(x)+f(x)=∫02​f(t)dt. If f(0)=e−2, then 2f(0)−f(2) is equal to ______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If dydx​=y1+x−y2​,x(1)=1, then 5x(2) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x=x(t) is the solution of the differential equation (t+1)dx=(2x+(t+1)4)dt,x(0)=2, then, x(1) equals

.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: If $x ϕ (x) = \int_{5}^{x} (3 t^{2} - 2 ϕ^{'} (t)) d t, x > - 2$ , and $ϕ (0) = 4$ then $ϕ (2)$ is.

Finding the Constant $C$

Finding the Expression for $ϕ (x)$

Let $F : [3, 5] \to R$ be a twice differentiable function on $(3, 5)$ such that $F (x) = e^{- x} \int_{3}^{x} (3 t^{2} + 2 t + 4 F^{'} (t)) d t$ . If $F^{'} (4) = \frac{α e ^{β} - 224}{( e ^{β} - 4 ) ^{2}}$ , then $α + β$ is equal to

Let $f : [1, \infty) \to [2, \infty)$ be a differentiable function such that $f (1) = 2$ . If $6 \int_{1}^{x} f (t) d t = 3 x f (x) - x^{3}$ for all $x \geq 1$ , then the value of $f (2)$ is .........

Let f be a differentiable function such that $2 (x + 2)^{2} f (x) - 3 (x + 2)^{2} = 10 \int_{0}^{x} (t + 2) f (t) d t, x \geq 0$ Then $f (2)$ 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

Let $f : R \to R$ be a differentiable function such that $f^{'} (x) + f (x) = \int_{0}^{2} f (t) d t$ . If $f (0) = e^{- 2}$ , then $2 f (0) - f (2)$ is equal to ______.

If $\frac{d x}{d y} = \frac{1 + x - y ^{2}}{y}, x (1) = 1$ , then $5 x (2)$ is equal to :

If $x = x (t)$ is the solution of the differential equation $(t + 1) d x = (2 x + (t + 1)^{4}) d t, x (0) = 2$ , then, $x (1)$ equals

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