Solve $T'(x)=45-0.75cdot T(x)$, where $T(0)=75$.












1












$begingroup$


I need help with homework:




Let $T'(x)=45-0.75cdot T(x)$, where $T(0)=75$. Find $T(x)$.




Here is what I have tried:



Let $T(x)=y.$ Rewrite the equation
$$frac{mathrm dy}{mathrm dx}=45-0.75yimplies frac{mathrm dy}{45-0.75y}=mathrm dx$$
Integrate both sides
$$int frac{mathrm dy}{45-0.75y}=int mathrm dx$$
Calculate LHS
$$-frac43ln(45-0.75y)+c=x$$
Now isolate $y$. Divide by "$-frac43$"
$$ln(45-0.75y)+c=-frac34x$$
Raise both sides in "$e$"
$$45-0.75y+c=e^{-frac34x}$$
Subtract $45$ and $c$, divide by "$-0.75$"
$$T(x)=y=frac{e^{-3x/4}-45}{0.75}-c$$
$$T(0)=75implies c=16.33$$
$$T(x)=frac{e^{-3x/4}-45}{0.75}-16.33$$




I know that the true answer is $$T(x)=15e^{-0.75 x}+60$$



However I am unsure on how to get that result.




This is the integral calculator for $1/(45-0,75x)$






integration ordinary-differential-equations derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    Solve linear diff. equation with constant coefficients $T'+frac34 T=45$
    $endgroup$
    – Aleksas Domarkas
    Nov 24 '18 at 10:35
















1












$begingroup$


I need help with homework:




Let $T'(x)=45-0.75cdot T(x)$, where $T(0)=75$. Find $T(x)$.




Here is what I have tried:



Let $T(x)=y.$ Rewrite the equation
$$frac{mathrm dy}{mathrm dx}=45-0.75yimplies frac{mathrm dy}{45-0.75y}=mathrm dx$$
Integrate both sides
$$int frac{mathrm dy}{45-0.75y}=int mathrm dx$$
Calculate LHS
$$-frac43ln(45-0.75y)+c=x$$
Now isolate $y$. Divide by "$-frac43$"
$$ln(45-0.75y)+c=-frac34x$$
Raise both sides in "$e$"
$$45-0.75y+c=e^{-frac34x}$$
Subtract $45$ and $c$, divide by "$-0.75$"
$$T(x)=y=frac{e^{-3x/4}-45}{0.75}-c$$
$$T(0)=75implies c=16.33$$
$$T(x)=frac{e^{-3x/4}-45}{0.75}-16.33$$




I know that the true answer is $$T(x)=15e^{-0.75 x}+60$$



However I am unsure on how to get that result.




This is the integral calculator for $1/(45-0,75x)$






integration ordinary-differential-equations derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    Solve linear diff. equation with constant coefficients $T'+frac34 T=45$
    $endgroup$
    – Aleksas Domarkas
    Nov 24 '18 at 10:35














1












1








1





$begingroup$


I need help with homework:




Let $T'(x)=45-0.75cdot T(x)$, where $T(0)=75$. Find $T(x)$.




Here is what I have tried:



Let $T(x)=y.$ Rewrite the equation
$$frac{mathrm dy}{mathrm dx}=45-0.75yimplies frac{mathrm dy}{45-0.75y}=mathrm dx$$
Integrate both sides
$$int frac{mathrm dy}{45-0.75y}=int mathrm dx$$
Calculate LHS
$$-frac43ln(45-0.75y)+c=x$$
Now isolate $y$. Divide by "$-frac43$"
$$ln(45-0.75y)+c=-frac34x$$
Raise both sides in "$e$"
$$45-0.75y+c=e^{-frac34x}$$
Subtract $45$ and $c$, divide by "$-0.75$"
$$T(x)=y=frac{e^{-3x/4}-45}{0.75}-c$$
$$T(0)=75implies c=16.33$$
$$T(x)=frac{e^{-3x/4}-45}{0.75}-16.33$$




I know that the true answer is $$T(x)=15e^{-0.75 x}+60$$



However I am unsure on how to get that result.




This is the integral calculator for $1/(45-0,75x)$






integration ordinary-differential-equations derivatives







share|cite|improve this question











$endgroup$




I need help with homework:




Let $T'(x)=45-0.75cdot T(x)$, where $T(0)=75$. Find $T(x)$.




Here is what I have tried:



Let $T(x)=y.$ Rewrite the equation
$$frac{mathrm dy}{mathrm dx}=45-0.75yimplies frac{mathrm dy}{45-0.75y}=mathrm dx$$
Integrate both sides
$$int frac{mathrm dy}{45-0.75y}=int mathrm dx$$
Calculate LHS
$$-frac43ln(45-0.75y)+c=x$$
Now isolate $y$. Divide by "$-frac43$"
$$ln(45-0.75y)+c=-frac34x$$
Raise both sides in "$e$"
$$45-0.75y+c=e^{-frac34x}$$
Subtract $45$ and $c$, divide by "$-0.75$"
$$T(x)=y=frac{e^{-3x/4}-45}{0.75}-c$$
$$T(0)=75implies c=16.33$$
$$T(x)=frac{e^{-3x/4}-45}{0.75}-16.33$$




I know that the true answer is $$T(x)=15e^{-0.75 x}+60$$



However I am unsure on how to get that result.




This is the integral calculator for $1/(45-0,75x)$






integration ordinary-differential-equations derivatives




integration ordinary-differential-equations derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 '18 at 12:30









LutzL

56.9k42054




56.9k42054










asked Nov 24 '18 at 10:16









Ryan CameronRyan Cameron

597




597












  • $begingroup$
    Solve linear diff. equation with constant coefficients $T'+frac34 T=45$
    $endgroup$
    – Aleksas Domarkas
    Nov 24 '18 at 10:35


















  • $begingroup$
    Solve linear diff. equation with constant coefficients $T'+frac34 T=45$
    $endgroup$
    – Aleksas Domarkas
    Nov 24 '18 at 10:35
















$begingroup$
Solve linear diff. equation with constant coefficients $T'+frac34 T=45$
$endgroup$
– Aleksas Domarkas
Nov 24 '18 at 10:35




$begingroup$
Solve linear diff. equation with constant coefficients $T'+frac34 T=45$
$endgroup$
– Aleksas Domarkas
Nov 24 '18 at 10:35










2 Answers
2






active

oldest

votes


















0












$begingroup$

I think you missed an absolute value (note that $45-0.75cdot 75<0$).



If we separate the variables, we have that
$$intfrac{dT}{45-0.75cdot T}=int 1dx$$
which gives
$$frac{ln(|45-0.75cdot T|)}{-0.75}=x+c.$$
By letting $T=75$ and $x=0$ we get $c=-ln(45/4)/0.75$.



Can you take it from here?



P.S. As suggested by Aleksas Domarkas, you may also solve the linear ODE with constant coefficients $T′+0.75T=45$ whose general solution is
$$T(x)=Ce^{-0.75 x}+60.$$
Now by using the initial value $T(0)=75$ to find $C=15$ and the solution is
$$T(x)=15e^{-0.75 x}+60.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Nov 24 '18 at 13:08



















0












$begingroup$

Your first grave error is that you used a wrong exponentiation rule. The exponential of $$ln(u)+c$$ is $$e^ccdot u$$ or $$Cu~~text{ with }~~C=pm e^c,$$ not $u+c$ or $u+e^c$ or whatever.



Changing the constant on the fly by incorporating other constant terms, which you do in practically every transformation while isolating $y$, is usually OK. But the number of changes should be kept to a minimum to avoid hard-to-reconstruct errors.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Well explained, thanks!
    $endgroup$
    – Ryan Cameron
    Nov 24 '18 at 12:38











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

I think you missed an absolute value (note that $45-0.75cdot 75<0$).



If we separate the variables, we have that
$$intfrac{dT}{45-0.75cdot T}=int 1dx$$
which gives
$$frac{ln(|45-0.75cdot T|)}{-0.75}=x+c.$$
By letting $T=75$ and $x=0$ we get $c=-ln(45/4)/0.75$.



Can you take it from here?



P.S. As suggested by Aleksas Domarkas, you may also solve the linear ODE with constant coefficients $T′+0.75T=45$ whose general solution is
$$T(x)=Ce^{-0.75 x}+60.$$
Now by using the initial value $T(0)=75$ to find $C=15$ and the solution is
$$T(x)=15e^{-0.75 x}+60.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Nov 24 '18 at 13:08
















0












$begingroup$

I think you missed an absolute value (note that $45-0.75cdot 75<0$).



If we separate the variables, we have that
$$intfrac{dT}{45-0.75cdot T}=int 1dx$$
which gives
$$frac{ln(|45-0.75cdot T|)}{-0.75}=x+c.$$
By letting $T=75$ and $x=0$ we get $c=-ln(45/4)/0.75$.



Can you take it from here?



P.S. As suggested by Aleksas Domarkas, you may also solve the linear ODE with constant coefficients $T′+0.75T=45$ whose general solution is
$$T(x)=Ce^{-0.75 x}+60.$$
Now by using the initial value $T(0)=75$ to find $C=15$ and the solution is
$$T(x)=15e^{-0.75 x}+60.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Nov 24 '18 at 13:08














0












0








0





$begingroup$

I think you missed an absolute value (note that $45-0.75cdot 75<0$).



If we separate the variables, we have that
$$intfrac{dT}{45-0.75cdot T}=int 1dx$$
which gives
$$frac{ln(|45-0.75cdot T|)}{-0.75}=x+c.$$
By letting $T=75$ and $x=0$ we get $c=-ln(45/4)/0.75$.



Can you take it from here?



P.S. As suggested by Aleksas Domarkas, you may also solve the linear ODE with constant coefficients $T′+0.75T=45$ whose general solution is
$$T(x)=Ce^{-0.75 x}+60.$$
Now by using the initial value $T(0)=75$ to find $C=15$ and the solution is
$$T(x)=15e^{-0.75 x}+60.$$






share|cite|improve this answer











$endgroup$



I think you missed an absolute value (note that $45-0.75cdot 75<0$).



If we separate the variables, we have that
$$intfrac{dT}{45-0.75cdot T}=int 1dx$$
which gives
$$frac{ln(|45-0.75cdot T|)}{-0.75}=x+c.$$
By letting $T=75$ and $x=0$ we get $c=-ln(45/4)/0.75$.



Can you take it from here?



P.S. As suggested by Aleksas Domarkas, you may also solve the linear ODE with constant coefficients $T′+0.75T=45$ whose general solution is
$$T(x)=Ce^{-0.75 x}+60.$$
Now by using the initial value $T(0)=75$ to find $C=15$ and the solution is
$$T(x)=15e^{-0.75 x}+60.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 24 '18 at 11:48

























answered Nov 24 '18 at 10:28









Robert ZRobert Z

94.5k1063134




94.5k1063134












  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Nov 24 '18 at 13:08


















  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – Aloizio Macedo♦
    Nov 24 '18 at 13:08
















$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Nov 24 '18 at 13:08




$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Nov 24 '18 at 13:08











0












$begingroup$

Your first grave error is that you used a wrong exponentiation rule. The exponential of $$ln(u)+c$$ is $$e^ccdot u$$ or $$Cu~~text{ with }~~C=pm e^c,$$ not $u+c$ or $u+e^c$ or whatever.



Changing the constant on the fly by incorporating other constant terms, which you do in practically every transformation while isolating $y$, is usually OK. But the number of changes should be kept to a minimum to avoid hard-to-reconstruct errors.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Well explained, thanks!
    $endgroup$
    – Ryan Cameron
    Nov 24 '18 at 12:38
















0












$begingroup$

Your first grave error is that you used a wrong exponentiation rule. The exponential of $$ln(u)+c$$ is $$e^ccdot u$$ or $$Cu~~text{ with }~~C=pm e^c,$$ not $u+c$ or $u+e^c$ or whatever.



Changing the constant on the fly by incorporating other constant terms, which you do in practically every transformation while isolating $y$, is usually OK. But the number of changes should be kept to a minimum to avoid hard-to-reconstruct errors.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Well explained, thanks!
    $endgroup$
    – Ryan Cameron
    Nov 24 '18 at 12:38














0












0








0





$begingroup$

Your first grave error is that you used a wrong exponentiation rule. The exponential of $$ln(u)+c$$ is $$e^ccdot u$$ or $$Cu~~text{ with }~~C=pm e^c,$$ not $u+c$ or $u+e^c$ or whatever.



Changing the constant on the fly by incorporating other constant terms, which you do in practically every transformation while isolating $y$, is usually OK. But the number of changes should be kept to a minimum to avoid hard-to-reconstruct errors.






share|cite|improve this answer









$endgroup$



Your first grave error is that you used a wrong exponentiation rule. The exponential of $$ln(u)+c$$ is $$e^ccdot u$$ or $$Cu~~text{ with }~~C=pm e^c,$$ not $u+c$ or $u+e^c$ or whatever.



Changing the constant on the fly by incorporating other constant terms, which you do in practically every transformation while isolating $y$, is usually OK. But the number of changes should be kept to a minimum to avoid hard-to-reconstruct errors.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 24 '18 at 12:36









LutzLLutzL

56.9k42054




56.9k42054












  • $begingroup$
    Well explained, thanks!
    $endgroup$
    – Ryan Cameron
    Nov 24 '18 at 12:38


















  • $begingroup$
    Well explained, thanks!
    $endgroup$
    – Ryan Cameron
    Nov 24 '18 at 12:38
















$begingroup$
Well explained, thanks!
$endgroup$
– Ryan Cameron
Nov 24 '18 at 12:38




$begingroup$
Well explained, thanks!
$endgroup$
– Ryan Cameron
Nov 24 '18 at 12:38


















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