Chapter 1: Q23E (page 1)

In Problems 21–26, solve the initial value problem.
$(e^{t} y + {te}^{t} y) dt + (t e^{t} + 2) dy = 0, y (0) = - 1$

Short Answer

Expert verified

The solution is . $y = \frac{- 2}{{te}^{t} + 2}$

Step by step solution

Evaluate the equation is exact

Here $(e^{t} y + {te}^{t} y) dt + (t e^{t} + 2) dy = 0, y (0) = - 1$

The condition for exact is $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}$ .

$M (t, y) = (e^{t} y + {te}^{t} y) N (t, y) = (t e^{t} + 2) \frac{\partial M}{\partial y} = e^{t} + {te}^{t} = \frac{\partial N}{\partial t}$

This equation is exact.

Find the value of F(x, y)

Here

$M (t, y) = (e^{t} y + {te}^{t} y) F (t, y) = \int M (t, y) dt + g (y) = \int (e^{t} y + t e^{t} y) dt + g (y) = y t e^{t} + g (y)$

Determine the value of g(y)

$\frac{\partial F}{\partial y} (t, y) = N (t, y) {te}^{t} + g' (y) = (t e^{t} + 2) g' (y) = 2 g (y) = 2 y + C_{1}$

Now

F (t, y) = yt e^{t} + 2 y + C_{1}

Therefore the solution of the differential equation is

${yte}^{t} + 2 y = C y = \frac{c}{{te}^{t} + 2}$

Apply the initial conditions.

$y (0) = - 1 y (0)$

$- 1 = \frac{c}{0 + 2} C = - 2$

The solutions is $y = \frac{- 2}{{te}^{t} + 2}$ .

Hence the solution is $y = \frac{- 2}{{te}^{t} + 2}$

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In Problems 21–26, solve the initial value problem.
$(e^{t} y + {te}^{t} y) dt + (t e^{t} + 2) dy = 0, y (0) = - 1$

Short Answer

Step by step solution

Evaluate the equation is exact

Find the value of F(x, y)

Determine the value of g(y)

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In Problems 21–26, solve the initial value problem.(ety+tety)dt+(tet+2)dy=0,y(0)=-1

Short Answer

Step by step solution

Evaluate the equation is exact

Find the value of F(x, y)

Determine the value of g(y)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Probability and Statistics

Mechanics Maths

Calculus

Discrete Mathematics

Applied Mathematics

Study anywhere. Anytime. Across all devices.

In Problems 21–26, solve the initial value problem.
$(e^{t} y + {te}^{t} y) dt + (t e^{t} + 2) dy = 0, y (0) = - 1$