Solved Obtain the general solution and particular solution | Chegg.com - Free Printable

Let's solve each of these separable differential equations step by step. We'll find the general solution using separation of variables, then apply the initial conditions to find the particular solution.

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1. $ yx^4 \frac{dy}{dx} = 1 $, with $ y = 0 $ when $ x = 1 $

#### Step 1: Separate variables
$$
yx^4 \frac{dy}{dx} = 1 \Rightarrow y\,dy = \frac{1}{x^4}\,dx
$$

#### Step 2: Integrate both sides
$$
\int y\,dy = \int x^{-4}\,dx
\Rightarrow \frac{y^2}{2} = \frac{x^{-3}}{-3} + C
\Rightarrow \frac{y^2}{2} = -\frac{1}{3x^3} + C
$$

Multiply both sides by 2:
$$
y^2 = -\frac{2}{3x^3} + 2C
\Rightarrow y^2 = -\frac{2}{3x^3} + C_1 \quad \text{(let } C_1 = 2C\text{)}
$$

#### Step 3: Apply initial condition $ y = 0 $ when $ x = 1 $
$$
0^2 = -\frac{2}{3(1)^3} + C_1 \Rightarrow 0 = -\frac{2}{3} + C_1 \Rightarrow C_1 = \frac{2}{3}
$$

So,
$$
y^2 = -\frac{2}{3x^3} + \frac{2}{3} = \frac{2}{3}\left(1 - \frac{1}{x^3}\right)
$$

Take square root:
$$
y = \pm \sqrt{ \frac{2}{3}\left(1 - \frac{1}{x^3}\right) }
$$

But at $ x = 1 $, $ y = 0 $. For $ x > 1 $, $ 1 - \frac{1}{x^3} > 0 $, so real solutions exist. However, we must check sign.

Since $ y = 0 $ at $ x = 1 $, and the expression is continuous, likely we take the positive or negative branch depending on behavior.

But let’s consider: from the ODE, $ yx^4 \frac{dy}{dx} = 1 $. At $ x = 1 $, $ y = 0 $, so left side is zero — but right side is 1 → contradiction?

Wait! This suggests a problem.

At $ x = 1 $, $ y = 0 $, then LHS = $ 0 \cdot (1)^4 \cdot \frac{dy}{dx} = 0 $, but RHS = 1 → $ 0 = 1 $? Contradiction!

So no solution exists satisfying $ y(1) = 0 $, because it violates the equation.

But wait — maybe I made an error in interpretation.

Wait: Let's double-check:

Given: $ yx^4 \frac{dy}{dx} = 1 $

If $ y = 0 $, then LHS = 0, but RHS = 1 → impossible.

So no solution exists that satisfies $ y(1) = 0 $.

But perhaps the initial condition is not consistent?

Wait — unless the initial condition is not on the solution curve.

But if $ y = 0 $ at $ x = 1 $, then plugging into ODE gives $ 0 = 1 $, which is false.

So the initial condition is inconsistent.

Thus, no solution exists for this initial value problem.

But maybe there's a typo? Or perhaps we should proceed formally.

Alternatively, maybe the initial condition is meant to be used after solving, but it leads to contradiction.

So no solution exists satisfying $ y(1) = 0 $.

But let’s suppose the initial condition is not contradictory — perhaps it's a mistake.

Wait — maybe the ODE is $ yx^4 \frac{dy}{dx} = 1 $, and $ y = 0 $ at $ x = 1 $, but that would require $ 0 = 1 $, impossible.

Hence, no solution exists.

But let’s assume instead that the initial condition is misstated, or perhaps it's a trick question.

Alternatively, maybe the ODE is $ yx^4 \frac{dy}{dx} = 1 $, and we solve generally, then see.

We have:
$$
y^2 = -\frac{2}{3x^3} + C_1
$$

For real $ y $, need $ -\frac{2}{3x^3} + C_1 \geq 0 $

But at $ x = 1 $, $ y = 0 $ ⇒ $ 0 = -\frac{2}{3} + C_1 $ ⇒ $ C_1 = \frac{2}{3} $

Then $ y^2 = \frac{2}{3}(1 - \frac{1}{x^3}) $

This is valid only when $ x^3 \geq 1 $, i.e., $ x \geq 1 $, since otherwise $ 1 - 1/x^3 < 0 $ → $ y^2 < 0 $, impossible.

So domain: $ x \geq 1 $

And at $ x = 1 $, $ y = 0 $

Now check if ODE holds at $ x = 1 $: $ y = 0 $, so $ yx^4 dy/dx = 0 \cdot 1 \cdot dy/dx = 0 $, but RHS = 1 → still contradiction.

So even though we get $ y = 0 $ at $ x = 1 $, the derivative blows up?

Let’s compute $ dy/dx $ from the solution.

From $ y^2 = \frac{2}{3}(1 - x^{-3}) $

Differentiate implicitly:
$$
2y \frac{dy}{dx} = \frac{2}{3} \cdot (3x^{-4}) = \frac{2}{x^4}
\Rightarrow \frac{dy}{dx} = \frac{1}{x^4 y}, \quad y \ne 0
$$

As $ x \to 1^+ $, $ y \to 0^+ $, so $ dy/dx \to \infty $

So $ yx^4 \frac{dy}{dx} = yx^4 \cdot \frac{1}{x^4 y} = 1 $ → holds for $ y \ne 0 $

But at $ y = 0 $, undefined.

So the solution approaches $ y = 0 $ as $ x \to 1^+ $, but never actually reaches it in finite derivative sense.

So the initial condition $ y(1) = 0 $ cannot be satisfied continuously unless we allow singularities.

Thus, no solution exists satisfying $ y(1) = 0 $.

But perhaps the problem has a typo. Maybe $ y = 1 $ at $ x = 1 $? But as stated, it's problematic.

Let’s move on and come back.

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2. $ y\,dx - x\,dy = 0 $, $ y = 10 $, $ x = 5 $

Rewrite:
$$
y\,dx = x\,dy \Rightarrow \frac{dy}{dx} = \frac{y}{x}
$$

Separate:
$$
\frac{dy}{y} = \frac{dx}{x}
$$

Integrate:
$$
\ln|y| = \ln|x| + C \Rightarrow |y| = C|x| \Rightarrow y = Cx
$$

Apply IC: $ y = 10 $, $ x = 5 $ → $ 10 = C \cdot 5 \Rightarrow C = 2 $

So particular solution: $ \boxed{y = 2x} $

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3. $ y^2 x^3 \frac{dy}{dx} = 2 $, $ y = 0 $, $ x = 1 $

Separate:
$$
y^2 dy = \frac{2}{x^3} dx
$$

Integrate:
$$
\int y^2 dy = \int 2x^{-3} dx
\Rightarrow \frac{y^3}{3} = 2 \cdot \frac{x^{-2}}{-2} + C = -x^{-2} + C
\Rightarrow \frac{y^3}{3} = -\frac{1}{x^2} + C
\Rightarrow y^3 = -\frac{3}{x^2} + 3C
\Rightarrow y^3 = -\frac{3}{x^2} + C_1
$$

Now apply $ y = 0 $ at $ x = 1 $:
$$
0 = -3 + C_1 \Rightarrow C_1 = 3
\Rightarrow y^3 = -\frac{3}{x^2} + 3 = 3\left(1 - \frac{1}{x^2}\right)
$$

So $ y = \sqrt[3]{3\left(1 - \frac{1}{x^2}\right)} $

But at $ x = 1 $, $ y = 0 $, good.

Check if ODE holds at $ x = 1 $: $ y = 0 $, so LHS: $ y^2 x^3 dy/dx = 0 $, RHS = 2 → again, $ 0 = 2 $? Contradiction.

Same issue as before.

So again, $ y = 0 $ at $ x = 1 $ makes LHS = 0, but RHS = 2 → impossible.

So no solution exists satisfying $ y(1) = 0 $.

Again, inconsistency.

But if we ignore that and just use the formula, we can write:

General solution: $ y^3 = -\frac{3}{x^2} + C $

Particular: $ y^3 = 3\left(1 - \frac{1}{x^2}\right) $

But at $ x = 1 $, $ y = 0 $, but derivative may blow up.

Compute $ dy/dx $:

From $ y^3 = 3 - 3x^{-2} $

Differentiate:
$$
3y^2 \frac{dy}{dx} = 6x^{-3} \Rightarrow \frac{dy}{dx} = \frac{2}{x^3 y^2}, \quad y \ne 0
$$

Then $ y^2 x^3 \frac{dy}{dx} = y^2 x^3 \cdot \frac{2}{x^3 y^2} = 2 $ → holds for $ y \ne 0 $

But at $ y = 0 $, undefined.

So again, solution approaches $ y = 0 $ as $ x \to 1 $, but doesn't satisfy the ODE at that point.

So initial condition not compatible.

But perhaps the problem allows it as a limit.

So we’ll write the solution as:
$$
y = \sqrt[3]{3\left(1 - \frac{1}{x^2}\right)}, \quad x > 1
$$

But strictly speaking, no solution satisfies $ y(1) = 0 $ due to contradiction.

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4. $ \frac{dy}{dx} = (1+x)y^3 $, $ y = \frac{1}{2} $, $ x = 0 $

Separate:
$$
\frac{dy}{y^3} = (1+x)\,dx
$$

Integrate:
$$
\int y^{-3} dy = \int (1+x) dx
\Rightarrow \frac{y^{-2}}{-2} = x + \frac{x^2}{2} + C
\Rightarrow -\frac{1}{2y^2} = x + \frac{x^2}{2} + C
$$

Multiply both sides by -2:
$$
\frac{1}{y^2} = -2x - x^2 - 2C
\Rightarrow \frac{1}{y^2} = -x^2 - 2x + C_1
$$

Apply IC: $ y = \frac{1}{2} $, $ x = 0 $

$ \frac{1}{(1/2)^2} = \frac{1}{1/4} = 4 $

So:
$$
4 = 0 + 0 + C_1 \Rightarrow C_1 = 4
\Rightarrow \frac{1}{y^2} = -x^2 - 2x + 4
\Rightarrow y^2 = \frac{1}{-x^2 - 2x + 4}
\Rightarrow y = \frac{1}{\sqrt{-x^2 - 2x + 4}}
$$

Domain: $ -x^2 - 2x + 4 > 0 $

Solve: $ x^2 + 2x - 4 < 0 $

Roots: $ x = \frac{-2 \pm \sqrt{4 + 16}}{2} = \frac{-2 \pm \sqrt{20}}{2} = -1 \pm \sqrt{5} $

So domain: $ x \in (-1 - \sqrt{5}, -1 + \sqrt{5}) $

At $ x = 0 $, within domain.

So particular solution:
$$
\boxed{y = \frac{1}{\sqrt{4 - x^2 - 2x}}}
$$

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5. $ \frac{dy}{dx} = (1+y)x^3 $, $ y = 0 $, $ x = 1 $

Separate:
$$
\frac{dy}{1+y} = x^3 dx
$$

Integrate:
$$
\ln|1+y| = \frac{x^4}{4} + C
\Rightarrow 1+y = Ce^{x^4/4}
\Rightarrow y = Ce^{x^4/4} - 1
$$

IC: $ y = 0 $, $ x = 1 $ → $ 0 = Ce^{1/4} - 1 \Rightarrow C = e^{-1/4} $

So:
$$
y = e^{-1/4} \cdot e^{x^4/4} - 1 = e^{(x^4 - 1)/4} - 1
$$

Answer:
$$
\boxed{y = e^{(x^4 - 1)/4} - 1}
$$

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6. $ \frac{dy}{dx} = \frac{\sin x}{\cos y} $, $ y = 0 $, $ x = 2\pi $

Separate:
$$
\cos y\,dy = \sin x\,dx
$$

Integrate:
$$
\int \cos y\,dy = \int \sin x\,dx
\Rightarrow \sin y = -\cos x + C
$$

IC: $ y = 0 $, $ x = 2\pi $

$ \sin 0 = 0 $, $ -\cos(2\pi) = -1 $, so:
$$
0 = -1 + C \Rightarrow C = 1
\Rightarrow \sin y = -\cos x + 1 = 1 - \cos x
$$

So:
$$
y = \arcsin(1 - \cos x)
$$

Note: $ 1 - \cos x \in [0,2] $, but $ \arcsin $ defined only for $ [-1,1] $, so $ 1 - \cos x \leq 1 $ → always true since $ \cos x \geq -1 $

But $ 1 - \cos x \geq 0 $, so $ y \geq 0 $

So:
$$
\boxed{y = \arcsin(1 - \cos x)}
$$

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7. $ \frac{dy}{dx} = \frac{4e^{3x}}{e^{3x}} = 4 $, $ y = 0 $, $ x = 0 $

Wait: $ \frac{dy}{dx} = \frac{4e^{3x}}{e^{3x}} = 4 $

So $ \frac{dy}{dx} = 4 $

Integrate: $ y = 4x + C $

IC: $ y = 0 $, $ x = 0 $ → $ 0 = 0 + C \Rightarrow C = 0 $

So:
$$
\boxed{y = 4x}
$$

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8. $ \frac{dy}{dx} = e^{2x - y} $, $ y = 0 $, $ x = 0 $

Write: $ \frac{dy}{dx} = e^{2x} e^{-y} $

Separate:
$$
e^y dy = e^{2x} dx
$$

Integrate:
$$
\int e^y dy = \int e^{2x} dx
\Rightarrow e^y = \frac{1}{2} e^{2x} + C
$$

Apply IC: $ y = 0 $, $ x = 0 $

$ e^0 = 1 $, $ \frac{1}{2} e^0 = \frac{1}{2} $, so:
$$
1 = \frac{1}{2} + C \Rightarrow C = \frac{1}{2}
\Rightarrow e^y = \frac{1}{2} e^{2x} + \frac{1}{2} = \frac{1}{2}(e^{2x} + 1)
$$

Take ln:
$$
y = \ln\left( \frac{e^{2x} + 1}{2} \right)
$$

So:
$$
\boxed{y = \ln\left( \frac{e^{2x} + 1}{2} \right)}
$$

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Summary of Solutions:

1. No solution exists due to inconsistency: $ y=0 $ at $ x=1 $ implies $ 0=1 $.
2. $ y = 2x $
3. No solution exists due to same reason.
4. $ y = \dfrac{1}{\sqrt{4 - x^2 - 2x}} $
5. $ y = e^{(x^4 - 1)/4} - 1 $
6. $ y = \arcsin(1 - \cos x) $
7. $ y = 4x $
8. $ y = \ln\left( \dfrac{e^{2x} + 1}{2} \right) $

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Note on Problems 1 and 3:

These have inconsistent initial conditions because substituting $ y = 0 $ into the ODE gives $ 0 = 1 $ or $ 0 = 2 $, which is impossible. So no solution exists satisfying those initial conditions.

However, if the initial condition were slightly different (e.g., $ y = \epsilon $), then a solution could be found approaching $ y = 0 $ asymptotically.

But as written, no solution exists for #1 and #3.

Let me know if you'd like to revise the initial conditions.