In many physical applications, the nonhomogeneous term F(x) is specified by different formulas in different intervals of x. (a) Find a general solution of the equation

Dottie Parra
2021-06-24
Answered

In many physical applications, the nonhomogeneous term F(x) is specified by different formulas in different intervals of x. (a) Find a general solution of the equation

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Luvottoq

Answered 2021-06-25
Author has **95** answers

a)$y=\{c1\mathrm{cos}x+{c}_{2}\mathrm{sin}x+x,0\le x\le 1;{c}_{1}\mathrm{cos}x+{c}_{2}\mathrm{sin}x+1,x\ge 1$

b)$y=\{x,0\le x\le 1;1,x\le 1$

b)

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Solve the differential equation by variation of parameters

$y"+y=\mathrm{sin}x$

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The integral surface of the first order partial differential equation

$2y(z-3)\frac{\mathrm{\partial}z}{\mathrm{\partial}x}+(2x-z)\frac{\mathrm{\partial}z}{\mathrm{\partial}y}=y(2x-3)$

passing through the curve ${x}^{2}+{y}^{2}=2x,z=0$ is

1. ${x}^{2}+{y}^{2}-{z}^{2}-2x+4z=0$

2. ${x}^{2}+{y}^{2}-{z}^{2}-2x+8z=0$

3. ${x}^{2}+{y}^{2}+{z}^{2}-2x+16z=0$

4. ${x}^{2}+{y}^{2}+{z}^{2}-2x+8z=0$

My effort:

I find the mulipliers x, 3y, -z and get the solution ${x}^{2}+3{y}^{2}-{z}^{2}={c}_{1}$. How to proceed further?

$2y(z-3)\frac{\mathrm{\partial}z}{\mathrm{\partial}x}+(2x-z)\frac{\mathrm{\partial}z}{\mathrm{\partial}y}=y(2x-3)$

passing through the curve ${x}^{2}+{y}^{2}=2x,z=0$ is

1. ${x}^{2}+{y}^{2}-{z}^{2}-2x+4z=0$

2. ${x}^{2}+{y}^{2}-{z}^{2}-2x+8z=0$

3. ${x}^{2}+{y}^{2}+{z}^{2}-2x+16z=0$

4. ${x}^{2}+{y}^{2}+{z}^{2}-2x+8z=0$

My effort:

I find the mulipliers x, 3y, -z and get the solution ${x}^{2}+3{y}^{2}-{z}^{2}={c}_{1}$. How to proceed further?

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Find the laplace transform of the following:

$a){t}^{2}\mathrm{sin}kt$

$b)t\mathrm{sin}kt$

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Let the first order differential equation be

$dy/dx+P(x)y=Q(x)$

To solve this we multiply it by function say v(x) then it becomes

$v(x)dy/dx+P(x)v(x)y=Q(x)v(x)$

$\therefore d(v(x)y)/dx=Q(x)v(x)$

where $d(v(x))/dx=P(x)v(x)$

I don't write full solution here but now here to obtain integrating factor v(x) from above equation we get $ln|v(x)|=\int P(x)dx$.

Now my question is why we neglect absolute value of v(x) and obtain integrating factor as ${e}^{\int P(x)dx}=v(x)$

$dy/dx+P(x)y=Q(x)$

To solve this we multiply it by function say v(x) then it becomes

$v(x)dy/dx+P(x)v(x)y=Q(x)v(x)$

$\therefore d(v(x)y)/dx=Q(x)v(x)$

where $d(v(x))/dx=P(x)v(x)$

I don't write full solution here but now here to obtain integrating factor v(x) from above equation we get $ln|v(x)|=\int P(x)dx$.

Now my question is why we neglect absolute value of v(x) and obtain integrating factor as ${e}^{\int P(x)dx}=v(x)$

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Using variable separable find the complete solution of the given differential equations

$x{y}^{3}dx+(y+1){e}^{-x}dy$

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My question is to find the solutions to the following

$\frac{df(x)}{dx}={f}^{-1}(x)$

where ${f}^{-1}(x)$ refers to the inverse of the function f. The domain really isn't important, though I am interested in either (-inf, inf) or (0, inf), so if any solutions are known for more restricted domains then they are welcome.

I cannot find any material relating to this type of question in any of my calculus and differential equations textbooks and references; it seems quite unorthodox. Any material which covers this type of diff equation would be wlecome

$\frac{df(x)}{dx}={f}^{-1}(x)$

where ${f}^{-1}(x)$ refers to the inverse of the function f. The domain really isn't important, though I am interested in either (-inf, inf) or (0, inf), so if any solutions are known for more restricted domains then they are welcome.

I cannot find any material relating to this type of question in any of my calculus and differential equations textbooks and references; it seems quite unorthodox. Any material which covers this type of diff equation would be wlecome