Solve $T'(x)=45-0.75cdot T(x)$, where $T(0)=75$.
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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 differential-equations derivatives
edited Nov 24 at 12:30
LutzL
55.2k42053
asked Nov 24 at 10:16
Ryan Cameron
407
add a comment |
up vote
1
down vote
favorite
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 differential-equations derivatives
edited Nov 24 at 12:30
LutzL
55.2k42053
asked Nov 24 at 10:16
Ryan Cameron
407
Solve linear diff. equation with constant coefficients $T'+frac34 T=45$
– Aleksas Domarkas
Nov 24 at 10:35
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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 differential-equations derivatives
edited Nov 24 at 12:30
LutzL
55.2k42053
asked Nov 24 at 10:16
Ryan Cameron
407
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 differential-equations derivatives
integration differential-equations derivatives
edited Nov 24 at 12:30
LutzL
55.2k42053
asked Nov 24 at 10:16
Ryan Cameron
407
edited Nov 24 at 12:30
LutzL
55.2k42053
asked Nov 24 at 10:16
Ryan Cameron
407
edited Nov 24 at 12:30
LutzL
55.2k42053
edited Nov 24 at 12:30
LutzL
55.2k42053
edited Nov 24 at 12:30
LutzL
55.2k42053
55.2k42053
asked Nov 24 at 10:16
Ryan Cameron
407
asked Nov 24 at 10:16
Ryan Cameron
407
asked Nov 24 at 10:16
Ryan Cameron
407
407
Solve linear diff. equation with constant coefficients $T'+frac34 T=45$
– Aleksas Domarkas
Nov 24 at 10:35
add a comment |
Solve linear diff. equation with constant coefficients $T'+frac34 T=45$
– Aleksas Domarkas
Nov 24 at 10:35
Solve linear diff. equation with constant coefficients $T'+frac34 T=45$
– Aleksas Domarkas
Nov 24 at 10:35
Solve linear diff. equation with constant coefficients $T'+frac34 T=45$
– Aleksas Domarkas
Nov 24 at 10:35
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
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
$$T(x)=15e^{-0.75 x}+60.$$
answered Nov 24 at 10:28
Robert Z
92.6k1060130
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Nov 24 at 13:08
add a comment |
up vote
0
down vote
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.
answered Nov 24 at 12:36
LutzL
55.2k42053
Well explained, thanks!
– Ryan Cameron
Nov 24 at 12:38
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
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
$$T(x)=15e^{-0.75 x}+60.$$
answered Nov 24 at 10:28
Robert Z
92.6k1060130
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Nov 24 at 13:08
add a comment |
up vote
0
down vote
accepted
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
$$T(x)=15e^{-0.75 x}+60.$$
answered Nov 24 at 10:28
Robert Z
92.6k1060130
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Nov 24 at 13:08
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
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
$$T(x)=15e^{-0.75 x}+60.$$
answered Nov 24 at 10:28
Robert Z
92.6k1060130
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
$$T(x)=15e^{-0.75 x}+60.$$
answered Nov 24 at 10:28
Robert Z
92.6k1060130
edited Nov 24 at 11:48
answered Nov 24 at 10:28
Robert Z
92.6k1060130
answered Nov 24 at 10:28
Robert Z
92.6k1060130
answered Nov 24 at 10:28
Robert Z
92.6k1060130
92.6k1060130
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Nov 24 at 13:08
add a comment |
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Nov 24 at 13:08
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Nov 24 at 13:08
Comments are not for extended discussion; this conversation has been moved to chat.
– Aloizio Macedo♦
Nov 24 at 13:08
add a comment |
up vote
0
down vote
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.
answered Nov 24 at 12:36
LutzL
55.2k42053
Well explained, thanks!
– Ryan Cameron
Nov 24 at 12:38
add a comment |
up vote
0
down vote
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.
answered Nov 24 at 12:36
LutzL
55.2k42053
Well explained, thanks!
– Ryan Cameron
Nov 24 at 12:38
add a comment |
up vote
0
down vote
up vote
0
down vote
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.
answered Nov 24 at 12:36
LutzL
55.2k42053
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.
answered Nov 24 at 12:36
LutzL
55.2k42053
answered Nov 24 at 12:36
LutzL
55.2k42053
answered Nov 24 at 12:36
LutzL
55.2k42053
answered Nov 24 at 12:36
LutzL
55.2k42053
55.2k42053
Well explained, thanks!
– Ryan Cameron
Nov 24 at 12:38
add a comment |
Well explained, thanks!
– Ryan Cameron
Nov 24 at 12:38
Well explained, thanks!
– Ryan Cameron
Nov 24 at 12:38
Well explained, thanks!
– Ryan Cameron
Nov 24 at 12:38
add a comment |
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Solve linear diff. equation with constant coefficients $T'+frac34 T=45$
– Aleksas Domarkas
Nov 24 at 10:35