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# решиние задачу

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y'+y=x

Guest Jun 29, 2017
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#1
+91247
+1

y'+y=x

My answer has been edited, thank you Alan for helping me fix it :)

\begin{align}\\ \frac{dy}{dx}+y&=x\\ (\frac{dy}{dx}+y)e^{x}&=xe^{x}\\ \frac{dy}{dx}e^{x}+ye^{x}&=xe^{x}\\ \text{This is the same as}\\ \frac{d}{dx}(ye^{x})&=xe^{x}\\ \int\frac{d}{dx}(ye^{x})\;dx&=\int xe^{x}\;dx\\ ye^{x}&=\int xe^{x}\;dx\\ \text{Integrate by parts}\\ \qquad u=x, \;\;v'=e^{x}\\ \qquad u'=1, \;\;v=e^{x}\\ ye^{x}&=\int xe^{x}\;dx\\ ye^{x}&=[uv]-\int vu'\;\;dx\\ ye^{x}&=[xe^{x}]-\int e^{x}\;\;dx\\ ye^{x}&=xe^{x}- e^{x}+k\\ y&=x- 1+\frac{k}{e^{x}}\\ y&=ke^{-x}+x-1\\ \end{align}\\

Melody  Jun 29, 2017
edited by Melody  Jun 30, 2017
#2
+18777
+1

y'+y=x

$$\begin{array}{|l|rcll|} \hline & y'+y &=& x \\ & y' &=& x -y \\ \text{substitution} \ u=x-y & y' &=& u \\ & 1-y' &=& 1-u \quad & | \quad u' = 1-y'\\ & u' &=& 1-u \\ & \frac{du}{dx} &=& 1-u \\ & \frac{du}{1-u} &=& dx \\ & \int \frac{du}{1-u} &=& \int dx \\ & -\ln|1-u| &=& x + c_1 \\ & \ln|1-u| &=& -x - c_1 \\ & 1-u &=& e^{-x - c_1} = c_2e^{-x} \quad & \quad (c_2=e^{-c_1}) \\ & u &=& 1- c_2e^{-x}=1-ce^{-x} \quad & \quad (c_2=c) \\ \text{back substitution} \ x-y=u & x-y &=& 1-ce^{-x} \\\\ & \mathbf{ y }& \mathbf{=} & \mathbf{ce^{-x} +x-1} \\ \hline \end{array}$$

proof y' + y = x

$$\begin{array}{|rcll|} \hline \mathbf{ y }& \mathbf{=} & \mathbf{ce^{-x} +x-1} \\ \mathbf{ y' }& \mathbf{=} & \mathbf{-ce^{-x} +1 } \\ y'+y &=& -ce^{-x} +1 + ce^{-x} +x-1 \\ y'+y &=& x \ \checkmark \\ \hline \end{array}$$

heureka  Jun 30, 2017
#3
+1

Solve the linear equation y(x) + ( dy(x))/( dx) = x:
Let μ(x) = e^( integral1 dx) = e^x.
Multiply both sides by μ(x):
e^x ( dy(x))/( dx) + e^x y(x) = e^x x
Substitute e^x = ( d)/( dx)(e^x):
e^x ( dy(x))/( dx) + ( d)/( dx)(e^x) y(x) = e^x x
Apply the reverse product rule g ( df)/( dx) + f ( dg)/( dx) = ( d)/( dx)(f g) to the left-hand side:
( d)/( dx)(e^x y(x)) = e^x x
Integrate both sides with respect to x:
integral( d)/( dx)(e^x y(x)) dx = integral e^x x dx
Evaluate the integrals:
e^x y(x) = e^x (x - 1) + c_1, where c_1 is an arbitrary constant.
Divide both sides by μ(x) = e^x:
Answer: | y(x) = x + c_1 e^(-x) - 1

Guest Jun 30, 2017
#4
+91247
0

Hi Guest,

Why don't you start learning LaTex?

It is not that hard, there is always people here that will help you get started.

You can start just by using it for fractions and very basic things and progress from there.

To write   $$5\frac{2}{3}$$

You just open the LaTex tab in the ribbon. Delete the formula that is already there and type in

5\frac{2}{3}          and press  ok :)

Melody  Jul 1, 2017
edited by Melody  Jul 1, 2017

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