Math Problem Statement

solve them by non separable method

Solution

Let's tackle the differential equations you provided using non-separable methods.

Problem 18

Given: dydx=1+ex+y+5\frac{dy}{dx} = 1 + e^{x+y+5}

Solution:

This differential equation is not separable in its current form. We'll use a substitution method to simplify it.

  1. Substitution:

    Let u=x+yu = x + y. Then, differentiate both sides with respect to xx: dudx=1+dydx\frac{du}{dx} = 1 + \frac{dy}{dx}

    Substitute this into the original equation: dudx=1+1+eu+5\frac{du}{dx} = 1 + 1 + e^{u+5} dudx=2+eu+5\frac{du}{dx} = 2 + e^{u+5}

  2. Simplification:

    This is a first-order linear differential equation in uu. To solve it, rewrite the equation: dudx2=eu+5\frac{du}{dx} - 2 = e^{u+5}

    To simplify further, let v=u+5v = u + 5 so that dvdx=dudx\frac{dv}{dx} = \frac{du}{dx} and the equation becomes: dvdx=2+ev\frac{dv}{dx} = 2 + e^{v}

  3. Solving the Equation:

    Now, rearrange and integrate: 12+evdv=dx\frac{1}{2 + e^{v}} dv = dx

    This integral may not have a simple analytical solution, but it can be approached by integrating numerically or using series expansions in certain cases. However, we can write the implicit solution as: dv2+ev=x+C\int \frac{dv}{2 + e^v} = x + C

    This expression represents the general solution, though it's not in a closed form.

Problem 20

Given: dydx=(1y4x1)2\frac{dy}{dx} = \left(\frac{1}{y - 4x - 1}\right)^2

Solution:

Again, this is not separable, but we can solve it using substitution.

  1. Substitution:

    Let u=y4x1u = y - 4x - 1. Then: dudx=dydx4\frac{du}{dx} = \frac{dy}{dx} - 4

    Substitute into the original equation: dudx=1u2+4\frac{du}{dx} = \frac{1}{u^2} + 4

  2. Simplification and Solution:

    The equation now becomes: dudx=4+1u2\frac{du}{dx} = 4 + \frac{1}{u^2}

    Multiply through by u2u^2 to obtain: u2dudx=4u2+1u^2 \frac{du}{dx} = 4u^2 + 1

    This can be written as: du4u2+1=dx\frac{du}{4u^2 + 1} = dx

    Integrating both sides: du4u2+1=x+C\int \frac{du}{4u^2 + 1} = x + C

    The integral on the left side is standard and can be solved using the arctangent formula: 12arctan(2u1)=x+C\frac{1}{2} \arctan\left(\frac{2u}{1}\right) = x + C

    Solving for uu: u=12tan(2x+2C)u = \frac{1}{2} \tan(2x + 2C)

  3. Back Substitution:

    Substituting back u=y4x1u = y - 4x - 1: y4x1=12tan(2x+2C)y - 4x - 1 = \frac{1}{2} \tan(2x + 2C)

    Finally, the general solution is: y=4x+1+12tan(2x+2C)y = 4x + 1 + \frac{1}{2} \tan(2x + 2C)

This represents the solution to the given differential equation.

Summary:

  • Substitution is a powerful technique for solving non-separable differential equations, especially when a direct separation of variables is impossible.
  • These equations often lead to integrals that can be solved using standard methods like arctangent formulas.

Would you like further details on any of the steps, or perhaps another example solved? Here are some related questions for further exploration:

  1. Solve dydx=y2+x2\frac{dy}{dx} = y^2 + x^2 using a suitable method.
  2. How do you solve a Riccati differential equation?
  3. Apply the method of undetermined coefficients to solve a non-homogeneous linear differential equation.
  4. What are the steps to solve an exact differential equation?
  5. Solve dydx=ex+y\frac{dy}{dx} = e^{x+y} using substitution.

Tip: When encountering complex differential equations, look for substitutions that simplify the problem into a form that you can handle, such as a linear equation or an equation with standard integrals.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Substitution Method
Integration Techniques

Formulas

Integration by parts
Arctangent formula

Theorems

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Suitable Grade Level

Advanced Mathematics