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Animated Solution for Mathematics - Differential Equations: If $\frac{d y}{d x} = y + 3 > 0$ and $y (0) = 2$ , then $y (ln 2)$ is equal to :

Q: Wait, why can't I just integrate $y+3$ directly with respect to $x$?

That is a common trap! You cannot integrate $y+3$ with respect to $x$ because $y$ is not a constant; it is a function of $x$. You must first separate the variables so that each side contains only one variable, allowing you to integrate with respect to that specific variable.

Q: Do I really need the constant $C$?

Absolutely. Without $C$, you have a general solution representing a family of curves. The initial condition $y(0) = 2$ acts as a 'key' that selects the one specific curve from that family that passes through the point $(0, 2)$.

Q: Is there a faster way using the linear differential equation formula?

Yes, you could rewrite it as $\frac{dy}{dx} - y = 3$ and use an integrating factor $e^{\int -1 dx} = e^{-x}$. It is a valid method, but for this specific problem, the variable separable method is much faster and less prone to calculation errors.

Select Answer:

Visualized Solution

The Differential Equation

Given Equation: $\frac{d y}{d x} = y + 3$
Constraint: $y + 3 > 0$
Initial Condition: $y (0) = 2$

Variable Separable Method

The equation is of the form $\frac{d y}{d x} = f (y)$ .
We can separate the variables $x$ and $y$ .

Separating the Variables

Rearrange the terms:
$\frac{d y}{y + 3} = d x$

Integrating Both Sides

Integrate both sides: $\int \frac{d y}{y + 3} = \int d x$
Result: $ln (y + 3) = x + C$

Applying Initial Condition

Use $y (0) = 2$ to find $C$ :
Substitute $x = 0$ and $y = 2$ :
$ln (2 + 3) = 0 + C$

Finding the Constant $C$

Simplify the equation:
$ln (5) = C$
Specific Solution: $ln (y + 3) = x + ln 5$

Setting up for $y (ln 2)$

We need to find $y$ when $x = ln 2$ .
Substitute $x = ln 2$ into the specific solution:
$ln (y + 3) = ln 2 + ln 5$

Using Log Properties

Use the property: $ln a + ln b = ln (ab)$
$ln (y + 3) = ln (2 \times 5)$
$ln (y + 3) = ln 10$

Final Calculation

Since $ln (y + 3) = ln 10$ , we can equate the arguments:
$y + 3 = 10$
$y = 7$
Final Answer: $y (ln 2) = 7$

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The Sigma Insight: Variable Separable Method

Solution Diagram

Analyzing the Setup

We are tasked with solving the differential equation:

\frac{d y}{d x} = y + 3

We are given the constraint

y + 3 > 0

and the initial condition

y (0) = 2

. Our objective is to determine the value of

y

when

x = ln 2

The Art of Separation

The first step is to recognize that the right-hand side depends entirely on

y

. This allows us to use the variable separable method.

By dividing both sides by

(y + 3)

and multiplying by

d x

, we obtain the symmetric equation:

\frac{d y}{y + 3} = d x

The variables are now perfectly separated and ready for integration.

The Constant of Destiny

Next, we integrate both sides of the equation:

\int \frac{d y}{y + 3} = \int d x

Performing the integration yields:

ln (y + 3) = x + C

Here,

C

is the constant of integration, representing the family of possible solutions. To find the specific curve for our problem, we apply the initial condition

y (0) = 2

Substituting

x = 0

and

y = 2

into the equation:

ln (2 + 3) = 0 + C \Rightarrow C = ln 5

Our specific solution is therefore:

ln (y + 3) = x + ln 5

The Logarithmic Climax

We now find the value of

y

when

x = ln 2

. Substituting this into our specific solution gives:

ln (y + 3) = ln 2 + ln 5

Using the logarithmic property

ln a + ln b = ln (ab)

, the right side simplifies:

ln (y + 3) = ln (2 \times 5) = ln 10

Since the natural logarithm is a one-to-one function, we equate the arguments:

y + 3 = 10

Solving for

y

, we arrive at the final result:

$y = 7$

Common Traps & FAQs

Wait, why can't I just integrate

y + 3

directly with respect to

x

That is a common trap! You cannot integrate

y + 3

with respect to

x

because

y

is not a constant; it is a function of

x

. You must first separate the variables so that each side contains only one variable, allowing you to integrate with respect to that specific variable.

Do I really need the constant

C

Absolutely. Without

C

, you have a general solution representing a family of curves. The initial condition

y (0) = 2

acts as a 'key' that selects the one specific curve from that family that passes through the point

(0, 2)

Is there a faster way using the linear differential equation formula?

Yes, you could rewrite it as

\frac{d y}{d x} - y = 3

and use an integrating factor

e^{\int - 1 d x} = e^{- x}

. It is a valid method, but for this specific problem, the variable separable method is much faster and less prone to calculation errors.

Animated Solution for Mathematics - Differential Equations: If $\frac{d y}{d x} = y + 3 > 0$ and $y (0) = 2$ , then $y (ln 2)$ is equal to :

Select Answer:

Visualized Solution

The Differential Equation

Variable Separable Method

Separating the Variables

Integrating Both Sides

Applying Initial Condition

Finding the Constant $C$

Setting up for $y (ln 2)$

Using Log Properties

Final Calculation

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Separation

The Constant of Destiny

The Logarithmic Climax

Similar Questions

If $y = y (x)$ is the solution of the differential equation, $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to

If $y = y (x)$ is the solution of the differential equation $(1 + e^{2 x}) \frac{d y}{d x} + 2 (1 + y^{2}) e^{x} = 0$ and $y (0) = 0$ , then $6 [y^{'} (0) + (y (lo g_{e} 3))^{2}]$ is equal to:

If $\frac{d y}{d x} = \frac{2 ^{x + y} - 2 ^{x}}{2 ^{y}}, y (0) = 1$ , then $y (1)$ is equal to :

If $\frac{d y}{d x} + \frac{2 ^{x - y} ( 2 ^{y} - 1 )}{2 ^{x} - 1} = 0, x, y > 0, y (1) = 1$ , then $y (2)$ is equal to :

Let for some function $y = f (x)$ , $\int_{0}^{x} t f (t) d t = x^{2} f (x)$ , $x > 0$ and $f (2) = 3$ Then $f (6)$ is equal to

If $y = y (x)$ is the solution of the differential equation $\frac{5 + e ^{x}}{2 + y} \frac{d y}{d x} + e^{x} = 0$ satisfying $y (0) = 1$ then a value of $y (lo g_{e} 13)$ is :

If $(2 + sin x) \frac{d y}{d x} + (y + 1) cos x = 0$ and $y (0) = 1$ , then $y (\frac{π}{2})$ is equal to:

Let $\int_{0}^{x} 1 - (y^{'} (t))^{2} d t = \int_{0}^{x} y (t) d t, 0 \leq x \leq 3, y \geq 0, y (0) = 0$ . Then at $x = 2, y^{''} + y + 1$ is equal to

Let $y = y (x)$ be the solution of the differential equation $d y = e^{α x + y} d x; α \in N$ . If $y (lo g_{e} 2) = lo g_{e} 2$ and $y (0) = lo g_{e} (1/2)$ , then the value of $α$ is equal to

If $x^{3} d y + x y d x = x^{2} d y + 2 y d x$ ; $y (2) = e$ and $x > 1$ , then $y (4)$ 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; } If dxdy​=y+3>0 and y(0)=2, then y(ln2) is equal to :

Select Answer:

Visualized Solution

The Differential Equation

Variable Separable Method

Separating the Variables

Integrating Both Sides

Applying Initial Condition

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

Setting up for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(ln2)

Using Log Properties

Final Calculation

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Separation

The Constant of Destiny

The Logarithmic Climax

Similar Questions

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.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 (1+e2x)dxdy​+2(1+y2)ex=0 and y(0)=0, then 6[y′(0)+(y(loge​3​))2] is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If dxdy​=2y2x+y−2x​,y(0)=1, then y(1) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If dxdy​+2x−12x−y(2y−1)​=0,x,y>0,y(1)=1, then y(2) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let for some function y=f(x), ∫0x​tf(t)dt=x2f(x), x>0 and f(2)=3 Then f(6) 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 2+y5+ex​dxdy​+ex=0 satisfying y(0)=1 then a value of y(loge​13) is :

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let ∫0x​1−(y′(t))2​dt=∫0x​y(t)dt,0≤x≤3,y≥0,y(0)=0. Then at x=2,y′′+y+1 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 dy=eαx+ydx;α∈N. If y(loge​2)=loge​2 and y(0)=loge​(1/2), then the value of α is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x3dy+xydx=x2dy+2ydx; y(2)=e and x>1, then y(4) is equal to :

Animated Solution for Mathematics - Differential Equations: If $\frac{d y}{d x} = y + 3 > 0$ and $y (0) = 2$ , then $y (ln 2)$ is equal to :

Finding the Constant $C$

Setting up for $y (ln 2)$

If $y = y (x)$ is the solution of the differential equation, $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to

If $y = y (x)$ is the solution of the differential equation $(1 + e^{2 x}) \frac{d y}{d x} + 2 (1 + y^{2}) e^{x} = 0$ and $y (0) = 0$ , then $6 [y^{'} (0) + (y (lo g_{e} 3))^{2}]$ is equal to:

If $\frac{d y}{d x} = \frac{2 ^{x + y} - 2 ^{x}}{2 ^{y}}, y (0) = 1$ , then $y (1)$ is equal to :

If $\frac{d y}{d x} + \frac{2 ^{x - y} ( 2 ^{y} - 1 )}{2 ^{x} - 1} = 0, x, y > 0, y (1) = 1$ , then $y (2)$ is equal to :

Let for some function $y = f (x)$ , $\int_{0}^{x} t f (t) d t = x^{2} f (x)$ , $x > 0$ and $f (2) = 3$ Then $f (6)$ is equal to

If $y = y (x)$ is the solution of the differential equation $\frac{5 + e ^{x}}{2 + y} \frac{d y}{d x} + e^{x} = 0$ satisfying $y (0) = 1$ then a value of $y (lo g_{e} 13)$ is :

If $(2 + sin x) \frac{d y}{d x} + (y + 1) cos x = 0$ and $y (0) = 1$ , then $y (\frac{π}{2})$ is equal to:

Let $\int_{0}^{x} 1 - (y^{'} (t))^{2} d t = \int_{0}^{x} y (t) d t, 0 \leq x \leq 3, y \geq 0, y (0) = 0$ . Then at $x = 2, y^{''} + y + 1$ is equal to

Let $y = y (x)$ be the solution of the differential equation $d y = e^{α x + y} d x; α \in N$ . If $y (lo g_{e} 2) = lo g_{e} 2$ and $y (0) = lo g_{e} (1/2)$ , then the value of $α$ is equal to

If $x^{3} d y + x y d x = x^{2} d y + 2 y d x$ ; $y (2) = e$ and $x > 1$ , then $y (4)$ is equal to :