The preparation process for differentiation strategies represents the process of differentiation under the polar coordinates, variables in equations. The differential equations may be present in the ordinary differential equations with different functions like algebraic functions, exponential functions, etc.. In this article we deal with the differential equations with the variables to differentiate for the strategy in the differentiation.
Preparation for Differentiation Strategies with Examples
Preparation for the differentiation strategies on the differential equations `[x+y]^2 dy/dx = 25`
Solution:
The equation given as `[x+y]^2 dy/dx = 25`
Put x + y = z
Differentiating with respect to 'x' we get,
`1 + dy/dx` = `dz/dx`
`=>` ` dy/dx` = `dz/dx -1`
The given equation becomes
`z^2[dz/dx - 1] = 25`
`=>` `dz/dx -1` = `25/z^2`
`=>` `dz/dx` = `1 + 25/z^2`
`=>` `dz/dx` = `[z^2+25]/z^2`
`=>` `dz` = `[z^2+25]/z^2 dx`
`=>` `dz``z^2/[z^2+25]` = `dx`
Integrating we have
`=>` `int z^2/[z^2+25]dz` = `int dx`
`=>` `int [z^2+25 - 25]/[z^2+25]dz` = `int dx`
`=>` `int [[z^2+25]/[z^2+25] - 25/[z^2+25]]dz` = `int dx`
`=>` `int [1 - 25/[z^2+25]]dz` = `int dx`
`=>` `int dz - int 25/[z^2+25]dz` = `int dx`
`=>` `z - 25 1/5 tan^-1z/5` = `x + c`
Put x + y = z
`=>` `x+y - 5. tan^-1[[x+y]/5]` = `x + c`
Reduce 'x' on both sides,
`=>` `y - 5. tan^-1[[x+y]/5]` = ` c` is the required solution.
Problems for the Preparation for Differentiation Strategies
Homework or practice problems on the preparation for differentiation strategies:
Solve `x dy = ( y + 4x^5 e^[x^4]) dx`
Solution:
The answer is `y/x = e^[x^4] + c`
Solve `[x^2 -y] dy + [y^2-x]dy =0`
Solution:
The answer is `x^3 + y^3 = 3xy`
Preparation for Differentiation Strategies with Examples
Preparation for the differentiation strategies on the differential equations `[x+y]^2 dy/dx = 25`
Solution:
The equation given as `[x+y]^2 dy/dx = 25`
Put x + y = z
Differentiating with respect to 'x' we get,
`1 + dy/dx` = `dz/dx`
`=>` ` dy/dx` = `dz/dx -1`
The given equation becomes
`z^2[dz/dx - 1] = 25`
`=>` `dz/dx -1` = `25/z^2`
`=>` `dz/dx` = `1 + 25/z^2`
`=>` `dz/dx` = `[z^2+25]/z^2`
`=>` `dz` = `[z^2+25]/z^2 dx`
`=>` `dz``z^2/[z^2+25]` = `dx`
Integrating we have
`=>` `int z^2/[z^2+25]dz` = `int dx`
`=>` `int [z^2+25 - 25]/[z^2+25]dz` = `int dx`
`=>` `int [[z^2+25]/[z^2+25] - 25/[z^2+25]]dz` = `int dx`
`=>` `int [1 - 25/[z^2+25]]dz` = `int dx`
`=>` `int dz - int 25/[z^2+25]dz` = `int dx`
`=>` `z - 25 1/5 tan^-1z/5` = `x + c`
Put x + y = z
`=>` `x+y - 5. tan^-1[[x+y]/5]` = `x + c`
Reduce 'x' on both sides,
`=>` `y - 5. tan^-1[[x+y]/5]` = ` c` is the required solution.
Problems for the Preparation for Differentiation Strategies
Homework or practice problems on the preparation for differentiation strategies:
Solve `x dy = ( y + 4x^5 e^[x^4]) dx`
Solution:
The answer is `y/x = e^[x^4] + c`
Solve `[x^2 -y] dy + [y^2-x]dy =0`
Solution:
The answer is `x^3 + y^3 = 3xy`
No comments:
Post a Comment