Determine the equation of the curve...

MathematicsVariable Separable MethodJEE Advanced 1996Moderate

Determine the equation of the curve passing through the origin, in the form $y = f (x)$ , which satisfies the differential equation $\frac{d y}{d x} = sin (10 x + 6 y)$ .

Answer:

y = \frac{1}{3} [tan^{- 1} (\frac{5 t a n 4 x}{4 - 3 t a n 4 x}) - 5 x]

Visualized Solution (Hindi)

Identifying the Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \frac{d y}{d x} = sin (10 x + 6 y)

Given differential equation: $\frac{d y}{d x} = sin (10 x + 6 y)$
Initial condition: The curve passes through the origin $(0, 0)$
Goal: Find the equation of the curve in the form $y = f (x)$

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v = 10 x + 6 y

Let $v = 10 x + 6 y$
Differentiating both sides with respect to $x$ :
$\frac{d v}{d x} = 10 + 6 \frac{d y}{d x}$

Transforming the Differential Equation

Substitute $\frac{d y}{d x} = sin v$ into the differentiated expression:
$\frac{d v}{d x} = 10 + 6 sin v$

Separating the Variables

Rearrange to separate variables $v$ and $x$ :
$\frac{d v}{10 + 6 s i n v} = d x$
Integrate both sides:
$\int \frac{d v}{10 + 6 s i n v} = \int d x$

Using Half-Angle Substitution

Use the identity: $sin v = \frac{2 t a n ( \frac{v}{2} )}{1 + t a n ^{2} ( \frac{v}{2} )}$
Let $t = tan (\frac{v}{2})$ , then $d v = \frac{2 d t}{1 + t ^{2}}$
The integral becomes: $\int \frac{\frac{2 d t}{1 + t ^{2}}}{10 + 6 ( \frac{2 t}{1 + t ^{2}} )} = \int d x$

Simplifying the Integral

Simplify the denominator: $10 (1 + t^{2}) + 12 t = 10 t^{2} + 12 t + 10$
The equation becomes: $\int \frac{2 d t}{10 t ^{2} + 12 t + 10} = x + C$
Divide by $2$ : $\int \frac{d t}{5 t ^{2} + 6 t + 5} = x + C$

Evaluating the Integral

Complete the square: $5 t^{2} + 6 t + 5 = 5 [(t + \frac{3}{5})^{2} + (\frac{4}{5})^{2}]$
Integral: $\frac{1}{5} \int \frac{d t}{( t + \frac{3}{5} ) ^{2} + ( \frac{4}{5} ) ^{2}} = x + C$
Using $\int \frac{d u}{u ^{2} + a ^{2}} = \frac{1}{a} tan^{- 1} (\frac{u}{a})$ :
$\frac{1}{5} \cdot \frac{5}{4} tan^{- 1} (\frac{t + \frac{3}{5}}{\frac{4}{5}}) = x + C \Rightarrow \frac{1}{4} tan^{- 1} (\frac{5 t + 3}{4}) = x + C$

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

At $(0, 0)$ , $x = 0$ and $y = 0 \Rightarrow v = 0 \Rightarrow t = tan (0) = 0$
Substitute into the equation: $\frac{1}{4} tan^{- 1} (\frac{5 ( 0 ) + 3}{4}) = 0 + C$
$C = \frac{1}{4} tan^{- 1} (\frac{3}{4})$

Simplifying with Inverse Trig Identities

Equation: $\frac{1}{4} tan^{- 1} (\frac{5 t + 3}{4}) - \frac{1}{4} tan^{- 1} (\frac{3}{4}) = x$
Multiply by $4$ : $tan^{- 1} (\frac{5 t + 3}{4}) - tan^{- 1} (\frac{3}{4}) = 4 x$
Apply $tan^{- 1} A - tan^{- 1} B = tan^{- 1} (\frac{A - B}{1 + A B})$ :
$tan^{- 1} (\frac{\frac{5 t + 3}{4} - \frac{3}{4}}{1 + ( \frac{5 t + 3}{4} ) ( \frac{3}{4} )}) = 4 x$

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Simplify the fraction: $\frac{5 t /4}{( 16 + 15 t + 9 ) /16} = \frac{5 t /4}{( 25 + 15 t ) /16} = \frac{4 t}{5 + 3 t}$
So, $tan^{- 1} (\frac{4 t}{5 + 3 t}) = 4 x \Rightarrow \frac{4 t}{5 + 3 t} = tan 4 x$
Solve for $t$ : $4 t = 5 tan 4 x + 3 t tan 4 x$
$t (4 - 3 tan 4 x) = 5 tan 4 x \Rightarrow t = \frac{5 t a n 4 x}{4 - 3 t a n 4 x}$

Final Expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y = f (x)

Recall $v = 2 tan^{- 1} t$ and $v = 10 x + 6 y$ :
$10 x + 6 y = 2 tan^{- 1} (\frac{5 t a n 4 x}{4 - 3 t a n 4 x})$
$6 y = 2 tan^{- 1} (\frac{5 t a n 4 x}{4 - 3 t a n 4 x}) - 10 x$
Divide by $6$ : $y = \frac{1}{3} [tan^{- 1} (\frac{5 t a n 4 x}{4 - 3 t a n 4 x}) - 5 x]$

Key Takeaways and Next Steps

Key Takeaway: Substitution $v = a x + b y + c$ reduces equations of form $\frac{d y}{d x} = f (a x + b y + c)$ to variable separable form.
Next Challenge: Try solving $\frac{d y}{d x} = cos (x + y)$ with the condition $y (0) = 0$ . How does the integration change?

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