1. ## does wolfram give wrong answer?

1. laplace transform unitstep(t-2)*(t-2)^2

laplace transform unitstep(t-2)*(t-2)^2 - Wolfram|Alpha

2.solving some system of linear differential equation also gives wrong answer

y'=y-x , x'=y+3x - Wolfram|Alpha

correct answer is y(t)= e^2t + te^2t + 0
x(t)= - e^2t - te^2t - e^2t

any ideas?

2. ## Re: does wolfram give wrong answer?

Originally Posted by kochibacha
1. laplace transform unitstep(t-2)*(t-2)^2

laplace transform unitstep(t-2)*(t-2)^2 - Wolfram|Alpha

2.solving some system of linear differential equation also gives wrong answer

y'=y-x , x'=y+3x - Wolfram|Alpha

correct answer is y(t)= e^2t + te^2t + 0
x(t)= - e^2t - te^2t - e^2t

any ideas?
Running the answer WA gives through the Simplify function yields the expression you have.

Your answer to the 2nd problem is incorrect. There are two constants of integration you've omitted.

3. ## Re: does wolfram give wrong answer?

Originally Posted by romsek
Running the answer WA gives through the Simplify function yields the expression you have.

Your answer to the 2nd problem is incorrect. There are two constants of integration you've omitted.

sorry , i forgot to tell where c1 and c2 = 1 in order to simplify the equation

even c1 and c2 are Arbitrary constant it still gives wrong answer

4. ## Re: does wolfram give wrong answer?

Originally Posted by kochibacha
1. laplace transform unitstep(t-2)*(t-2)^2

laplace transform unitstep(t-2)*(t-2)^2 - Wolfram|Alpha

2.solving some system of linear differential equation also gives wrong answer

y'=y-x , x'=y+3x - Wolfram|Alpha

correct answer is y(t)= e^2t + te^2t + 0
x(t)= - e^2t - te^2t - e^2t

any ideas?
\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}t} &= y - x \\ \frac{\mathrm{d}x}{\mathrm{d}t} &= y + 3x \\ \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{ \mathrm{d} x}{\mathrm{d}t}} \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{y - x}{y + 3x} \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\frac{1}{x} \left( y - x \right) }{\frac{1}{x} \left( y + 3x \right) } \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\frac{y}{x} - 1}{\frac{y}{x} + 3} \end{align*}

Now make the substitution \displaystyle \begin{align*} v = \frac{y}{x} \implies y = v\,x \implies \frac{\mathrm{d}y}{\mathrm{d}x} = v + x\,\frac{\mathrm{d}v}{\mathrm{d}x} \end{align*} and the DE becomes

\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\frac{y}{x} - 1}{\frac{y}{x} + 3} \\ v + x\,\frac{\mathrm{d}v}{\mathrm{d}x} &= \frac{v - 1}{v + 3} \\ x\,\frac{\mathrm{d}v}{\mathrm{d}x} &= \frac{v - 1}{v + 3} - v \\ x\,\frac{\mathrm{d}v}{\mathrm{d}x} &= \frac{v - 1}{v + 3} - \frac{v \left( v + 3 \right) }{ v + 3 } \\ x\,\frac{\mathrm{d}v}{\mathrm{d}x} &= \frac{-v^2 - 2v - 1}{v + 3} \\ \frac{v + 3}{v^2 + 2v + 1}\,\frac{\mathrm{d}v}{\mathrm{d}x} &= -\frac{1}{x} \\ \int{ \frac{v + 3}{ \left( v + 1 \right) ^2 } \, \frac{\mathrm{d}v}{\mathrm{d}x}\,\mathrm{d}x} &= \int{ -\frac{1}{x} \, \mathrm{d}x} \\ \int{ \frac{v + 3}{\left( v + 1 \right) ^2 } \,\mathrm{d}v } &= -\ln{ \left| x \right| } + C_1 \end{align*}

Now make the substitution \displaystyle \begin{align*} u = v + 1 \implies \mathrm{d}u = \mathrm{d}v \end{align*} and the integral becomes

\displaystyle \begin{align*} \int{ \frac{v + 1 + 2}{\left( v + 1 \right) ^2 } \,\mathrm{d}v} &= -\ln{ |x| } +C_1 \\ \int{ \frac{ u + 2}{u^2} \, \mathrm{d}u } &= -\ln{|x|} + C_1 \\ \int{ \frac{1}{u} + 2u^{-2} \, \mathrm{d}u} &= -\ln{|x|} + C_1 \\ \ln{|u|} - 2u^{-1} + C_2 &= -\ln{|x|} + C_1 \\ \ln{ | v + 1 | } - \frac{2}{v + 1} + C_2 &= -\ln{|x|} + C_1 \\ \ln{ \left| \frac{y}{x} + 1 \right| } - \frac{2}{\frac{y}{x} + 1 } + C_2 &= -\ln{|x|} + C_1 \\ \ln{ \left| \frac{y + x}{x} \right| } - \frac{2x}{y + x} + C_2 &= -\ln{|x|} + C_1 \\ \ln{ \left| \frac{y+x}{x} \right| } + \ln{|x|} &= \frac{2x}{y + x} + C_1 - C_2 \\ \ln{ \left| y + x \right| } &= \frac{2x}{y + x} + C_2 - C_1 \\ \left| y + x \right| &= \mathrm{e}^{\frac{2x}{y+x} + C_2 - C_1 } \\ |y + x| &= \mathrm{e}^{C_2 - C_1} \, \mathrm{e}^{\frac{2x}{y + x}} \\ y + x &= C\,\mathrm{e}^{\frac{2x}{y + x}} \textrm{ where } C = \pm \mathrm{e}^{C_2 - C_1} \end{align*}

5. ## Re: does wolfram give wrong answer?

Originally Posted by Prove It
\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}t} &= y - x \\ \frac{\mathrm{d}x}{\mathrm{d}t} &= y + 3x \\ \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{ \mathrm{d} x}{\mathrm{d}t}} \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{y - x}{y + 3x} \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\frac{1}{x} \left( y - x \right) }{\frac{1}{x} \left( y + 3x \right) } \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\frac{y}{x} - 1}{\frac{y}{x} + 3} \end{align*}

Now make the substitution \displaystyle \begin{align*} v = \frac{y}{x} \implies y = v\,x \implies \frac{\mathrm{d}y}{\mathrm{d}x} = v + x\,\frac{\mathrm{d}v}{\mathrm{d}x} \end{align*} and the DE becomes

\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\frac{y}{x} - 1}{\frac{y}{x} + 3} \\ v + x\,\frac{\mathrm{d}v}{\mathrm{d}x} &= \frac{v - 1}{v + 3} \\ x\,\frac{\mathrm{d}v}{\mathrm{d}x} &= \frac{v - 1}{v + 3} - v \\ x\,\frac{\mathrm{d}v}{\mathrm{d}x} &= \frac{v - 1}{v + 3} - \frac{v \left( v + 3 \right) }{ v + 3 } \\ x\,\frac{\mathrm{d}v}{\mathrm{d}x} &= \frac{-v^2 - 2v - 1}{v + 3} \\ \frac{v + 3}{v^2 + 2v + 1}\,\frac{\mathrm{d}v}{\mathrm{d}x} &= -\frac{1}{x} \\ \int{ \frac{v + 3}{ \left( v + 1 \right) ^2 } \, \frac{\mathrm{d}v}{\mathrm{d}x}\,\mathrm{d}x} &= \int{ -\frac{1}{x} \, \mathrm{d}x} \\ \int{ \frac{v + 3}{\left( v + 1 \right) ^2 } \,\mathrm{d}v } &= -\ln{ \left| x \right| } + C_1 \end{align*}

Now make the substitution \displaystyle \begin{align*} u = v + 1 \implies \mathrm{d}u = \mathrm{d}v \end{align*} and the integral becomes

\displaystyle \begin{align*} \int{ \frac{v + 1 + 2}{\left( v + 1 \right) ^2 } \,\mathrm{d}v} &= -\ln{ |x| } +C_1 \\ \int{ \frac{ u + 2}{u^2} \, \mathrm{d}u } &= -\ln{|x|} + C_1 \\ \int{ \frac{1}{u} + 2u^{-2} \, \mathrm{d}u} &= -\ln{|x|} + C_1 \\ \ln{|u|} - 2u^{-1} + C_2 &= -\ln{|x|} + C_1 \\ \ln{ | v + 1 | } - \frac{2}{v + 1} + C_2 &= -\ln{|x|} + C_1 \\ \ln{ \left| \frac{y}{x} + 1 \right| } - \frac{2}{\frac{y}{x} + 1 } + C_2 &= -\ln{|x|} + C_1 \\ \ln{ \left| \frac{y + x}{x} \right| } - \frac{2x}{y + x} + C_2 &= -\ln{|x|} + C_1 \\ \ln{ \left| \frac{y+x}{x} \right| } + \ln{|x|} &= \frac{2x}{y + x} + C_1 - C_2 \\ \ln{ \left| y + x \right| } &= \frac{2x}{y + x} + C_2 - C_1 \\ \left| y + x \right| &= \mathrm{e}^{\frac{2x}{y+x} + C_2 - C_1 } \\ |y + x| &= \mathrm{e}^{C_2 - C_1} \, \mathrm{e}^{\frac{2x}{y + x}} \\ y + x &= C\,\mathrm{e}^{\frac{2x}{y + x}} \textrm{ where } C = \pm \mathrm{e}^{C_2 - C_1} \end{align*}
I understand nothing about this post, integration etc. But hats off for typing such a huge math with LaTex. How much time did you take?

6. ## Re: does wolfram give wrong answer?

You know that it's integration...

I find it very hard to believe that you are given a problem about solving a system of DEs and have no idea about how to solve a DE...

7. ## Re: does wolfram give wrong answer?

Originally Posted by Prove It
You know that it's integration...
I know it is integration because
Originally Posted by Prove It

Now make the substitution \displaystyle \begin{align*} u = v + 1 \implies \mathrm{d}u = \mathrm{d}v \end{align*} and the integral becomes

8. ## Re: does wolfram give wrong answer?

Originally Posted by NameIsHidden
I know it is integration because
My apologies, I mistook you as being the original poster.

It didn't take me all that long to type up really, just because I can touch type quickly and know the coding...