Chapter 2: Q19E (page 76)

Use the method discussed under “Equations of the Form $\frac{dy}{dx} = G (ax + by)$ ” to solve problems17-20.
$\frac{dy}{dx} = {(x - y + 5)}^{2}$

Short Answer

Expert verified

$y = x + \frac{(6 + 4 {Ce}^{2 x})}{(1 + {Ce}^{2 x})}$ and $y = x + 4$

Step by step solution

Evaluate the given equation

Given,

$\frac{dy}{dx} = {(x - y + 5)}^{2} . . . . . . (1)$

Let us take

$z = x - y + 5 \to y = x - z + 5$

Differentiate with respect tox.

$\frac{dy}{dx} = 1 - \frac{dz}{dx}$

Substitute in equation (1).

$\begin{array}{l} 1 - \frac{dz}{dx} = {(z + 5)}^{2} \\ \frac{dz}{dx} = 1 - {(z + 5)}^{2} \\ \frac{dz}{1 - {(z + 5)}^{2}} = dx \end{array}$

So, $\frac{dz}{1 - {(z + 5)}^{2}} = dx . . . . . (2)$

Integrate the equation

Now, integrate the equation (2) on both sides.

$\int \frac{1}{1 - {(z + 5)}^{2}} dz = \int dx$

Now using the integral property: $\int \frac{du}{a^{2} - u^{2}} = \frac{1}{2 a} \ln |\frac{a + u}{a - u}| + C$ . Find the value of y.

$\begin{matrix} \frac{1}{2} \ln |\frac{1 + (z + 5)}{1 - (z + 5)}| = x + C_{1} \\ \ln |\frac{1 + z + 5}{1 - z - 5}| = 2 x + C_{1} \\ \ln |\frac{z + 6}{- (z + 4)}| = 2 x + C_{1} \end{matrix}$

Substitution method

Take log on both sides.

$\begin{array}{l} \frac{z + 6}{- (z + 4)} = {Ce}^{2 x} \\ z + 6 = - (z + 4) {Ce}^{2 x} \\ = - {zCe}^{2 x} - 4 {Ce}^{2 x} \\ z + {zCe}^{2 x} = - 6 - 4 {Ce}^{2 x} \\ z (1 + {Ce}^{2 x}) = - (6 + 4 {Ce}^{2 x}) \\ z = \frac{- (6 + 4 {Ce}^{2 x})}{(1 + {Ce}^{2 x})} \end{array}$

Now substitute the value ofzhere.

$\begin{array}{l} x - y = \frac{- (6 + 4 {Ce}^{2 x})}{(1 + {Ce}^{2 x})} \\ - y = - x - \frac{6 + 4 {Ce}^{2 x}}{1 + {Ce}^{2 x}} \\ y = x + \frac{(6 + 4 {Ce}^{2 x})}{(1 + {Ce}^{2 x})} \end{array}$

And one more possible solution is $y = x + 4$ .

Hence, the solutions of the given initial value problem is $y = x + \frac{(6 + 4 {Ce}^{2 x})}{(1 + {Ce}^{2 x})}$ and y=x+4.

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Use the method discussed under “Equations of the Form $\frac{dy}{dx} = G (ax + by)$ ” to solve problems17-20.
$\frac{dy}{dx} = {(x - y + 5)}^{2}$

Short Answer

Step by step solution

Evaluate the given equation

Integrate the equation

Substitution method

One App. One Place for Learning.

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Use the method discussed under “Equations of the Form dydx=Gax+by” to solve problems17-20.dydx=x-y+52

Short Answer

Step by step solution

Evaluate the given equation

Integrate the equation

Substitution method

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Theoretical and Mathematical Physics

Pure Maths

Applied Mathematics

Decision Maths

Geometry

Study anywhere. Anytime. Across all devices.

Use the method discussed under “Equations of the Form $\frac{dy}{dx} = G (ax + by)$ ” to solve problems17-20.
$\frac{dy}{dx} = {(x - y + 5)}^{2}$