Mixing
Differential Equations Solving for y
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I was wondering whether anyone could help me with a question which I am currently stuck on? It is asking for the general solution for the expression.
$$frac dydx=frac 1 y-3$$
$$y > 3 ,y(2)=5$$
I have Reciprocated both sides, establishing that
$$x = int (y-3)dy$$
but then it forms a quadratic and I cant seem to find the expression for y. It is probaby really easy I just cant seem to do it, thanks for your help!
differential-equations
edited 2 days ago
Isham
10.4k3829
asked 2 days ago
Joshua Peters
93
add a comment |Â
up vote
1
down vote
favorite
I was wondering whether anyone could help me with a question which I am currently stuck on? It is asking for the general solution for the expression.
$$frac dydx=frac 1 y-3$$
$$y > 3 ,y(2)=5$$
I have Reciprocated both sides, establishing that
$$x = int (y-3)dy$$
but then it forms a quadratic and I cant seem to find the expression for y. It is probaby really easy I just cant seem to do it, thanks for your help!
differential-equations
edited 2 days ago
Isham
10.4k3829
asked 2 days ago
Joshua Peters
93
Sometimes it is just not possible or worth the effort to get an explicit solution in terms of $y$
– Gobabis
2 days ago
What quadratic equation have you got? After seperating the variables the equation is $(y-3) dy=dxRightarrow int (y-3) dy=int dx$
– callculus
2 days ago
Have you solved the problem by using the hint? If not, give a reply. If yes, give a reply as well.
– callculus
2 days ago
@Gobabis It doesn´t take much effort to solve the equation.
– callculus
2 days ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I was wondering whether anyone could help me with a question which I am currently stuck on? It is asking for the general solution for the expression.
$$frac dydx=frac 1 y-3$$
$$y > 3 ,y(2)=5$$
I have Reciprocated both sides, establishing that
$$x = int (y-3)dy$$
but then it forms a quadratic and I cant seem to find the expression for y. It is probaby really easy I just cant seem to do it, thanks for your help!
differential-equations
edited 2 days ago
Isham
10.4k3829
asked 2 days ago
Joshua Peters
93
I was wondering whether anyone could help me with a question which I am currently stuck on? It is asking for the general solution for the expression.
$$frac dydx=frac 1 y-3$$
$$y > 3 ,y(2)=5$$
I have Reciprocated both sides, establishing that
$$x = int (y-3)dy$$
but then it forms a quadratic and I cant seem to find the expression for y. It is probaby really easy I just cant seem to do it, thanks for your help!
differential-equations
edited 2 days ago
Isham
10.4k3829
asked 2 days ago
Joshua Peters
93
edited 2 days ago
Isham
10.4k3829
edited 2 days ago
Isham
10.4k3829
edited 2 days ago
Isham
10.4k3829
10.4k3829
asked 2 days ago
Joshua Peters
93
asked 2 days ago
Joshua Peters
93
asked 2 days ago
Joshua Peters
93
93
Sometimes it is just not possible or worth the effort to get an explicit solution in terms of $y$
– Gobabis
2 days ago
What quadratic equation have you got? After seperating the variables the equation is $(y-3) dy=dxRightarrow int (y-3) dy=int dx$
– callculus
2 days ago
Have you solved the problem by using the hint? If not, give a reply. If yes, give a reply as well.
– callculus
2 days ago
@Gobabis It doesn´t take much effort to solve the equation.
– callculus
2 days ago
add a comment |Â
Sometimes it is just not possible or worth the effort to get an explicit solution in terms of $y$
– Gobabis
2 days ago
What quadratic equation have you got? After seperating the variables the equation is $(y-3) dy=dxRightarrow int (y-3) dy=int dx$
– callculus
2 days ago
Have you solved the problem by using the hint? If not, give a reply. If yes, give a reply as well.
– callculus
2 days ago
@Gobabis It doesn´t take much effort to solve the equation.
– callculus
2 days ago
Sometimes it is just not possible or worth the effort to get an explicit solution in terms of $y$
– Gobabis
2 days ago
Sometimes it is just not possible or worth the effort to get an explicit solution in terms of $y$
– Gobabis
2 days ago
What quadratic equation have you got? After seperating the variables the equation is $(y-3) dy=dxRightarrow int (y-3) dy=int dx$
– callculus
2 days ago
What quadratic equation have you got? After seperating the variables the equation is $(y-3) dy=dxRightarrow int (y-3) dy=int dx$
– callculus
2 days ago
Have you solved the problem by using the hint? If not, give a reply. If yes, give a reply as well.
– callculus
2 days ago
Have you solved the problem by using the hint? If not, give a reply. If yes, give a reply as well.
– callculus
2 days ago
@Gobabis It doesn´t take much effort to solve the equation.
– callculus
2 days ago
@Gobabis It doesn´t take much effort to solve the equation.
– callculus
2 days ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
$$frac dydx=frac 1 y-3$$
The differential equation is separable
$$int y-3 dy= int dx$$
$$frac y^22-3y=x+K$$
$$y^2-6y=2x+K$$
Complete the square
$$y^2-6y+colorred9=2x+K+colorred9$$
$$(y-3)^2=2x+C$$
$$y(2)=5 implies 2^2=4+C implies C=0$$
Therefore
$$(y-3)^2=2x$$
$$y(x)=3+ sqrt 2x$$
answered 2 days ago
Isham
10.4k3829
You have just integrated it as $int(y-3)dy = frac(y-3)^22=x+C$. No need to complete the square.
– Dylan
yesterday
add a comment |Â
up vote
0
down vote
Just a bit different approach:
$$fracmathrmdymathrmdx=frac1y-3$$
Let $u=y-3$, then we have that
$$fracmathrmdumathrmdx=fracmathrmdymathrmdx$$
So
$$fracmathrmdumathrmdx=frac1u$$
$$ufracmathrmdumathrmdx=1$$
$$2ufracmathrmdumathrmdx=2$$
$$fracmathrmdu^2mathrmdx=2$$
$$u^2=2x+C$$
And if we use the initial conditions now:
$$y(2)=u(2)+3=5$$
$$u(2)=2$$
So
$$4=4+C$$
$$C=0$$
Back to the equation:
$$u^2=2x$$
$$u=sqrt2x$$
$$y-3=sqrt2x$$
$$y(x)=sqrt2x+3$$
answered 2 days ago
Botond
3,8702532
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
$$frac dydx=frac 1 y-3$$
The differential equation is separable
$$int y-3 dy= int dx$$
$$frac y^22-3y=x+K$$
$$y^2-6y=2x+K$$
Complete the square
$$y^2-6y+colorred9=2x+K+colorred9$$
$$(y-3)^2=2x+C$$
$$y(2)=5 implies 2^2=4+C implies C=0$$
Therefore
$$(y-3)^2=2x$$
$$y(x)=3+ sqrt 2x$$
answered 2 days ago
Isham
10.4k3829
You have just integrated it as $int(y-3)dy = frac(y-3)^22=x+C$. No need to complete the square.
– Dylan
yesterday
add a comment |Â
up vote
2
down vote
$$frac dydx=frac 1 y-3$$
The differential equation is separable
$$int y-3 dy= int dx$$
$$frac y^22-3y=x+K$$
$$y^2-6y=2x+K$$
Complete the square
$$y^2-6y+colorred9=2x+K+colorred9$$
$$(y-3)^2=2x+C$$
$$y(2)=5 implies 2^2=4+C implies C=0$$
Therefore
$$(y-3)^2=2x$$
$$y(x)=3+ sqrt 2x$$
answered 2 days ago
Isham
10.4k3829
You have just integrated it as $int(y-3)dy = frac(y-3)^22=x+C$. No need to complete the square.
– Dylan
yesterday
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$$frac dydx=frac 1 y-3$$
The differential equation is separable
$$int y-3 dy= int dx$$
$$frac y^22-3y=x+K$$
$$y^2-6y=2x+K$$
Complete the square
$$y^2-6y+colorred9=2x+K+colorred9$$
$$(y-3)^2=2x+C$$
$$y(2)=5 implies 2^2=4+C implies C=0$$
Therefore
$$(y-3)^2=2x$$
$$y(x)=3+ sqrt 2x$$
answered 2 days ago
Isham
10.4k3829
$$frac dydx=frac 1 y-3$$
The differential equation is separable
$$int y-3 dy= int dx$$
$$frac y^22-3y=x+K$$
$$y^2-6y=2x+K$$
Complete the square
$$y^2-6y+colorred9=2x+K+colorred9$$
$$(y-3)^2=2x+C$$
$$y(2)=5 implies 2^2=4+C implies C=0$$
Therefore
$$(y-3)^2=2x$$
$$y(x)=3+ sqrt 2x$$
answered 2 days ago
Isham
10.4k3829
edited 2 days ago
answered 2 days ago
Isham
10.4k3829
answered 2 days ago
Isham
10.4k3829
answered 2 days ago
Isham
10.4k3829
10.4k3829
You have just integrated it as $int(y-3)dy = frac(y-3)^22=x+C$. No need to complete the square.
– Dylan
yesterday
add a comment |Â
You have just integrated it as $int(y-3)dy = frac(y-3)^22=x+C$. No need to complete the square.
– Dylan
yesterday
You have just integrated it as $int(y-3)dy = frac(y-3)^22=x+C$. No need to complete the square.
– Dylan
yesterday
You have just integrated it as $int(y-3)dy = frac(y-3)^22=x+C$. No need to complete the square.
– Dylan
yesterday
add a comment |Â
up vote
0
down vote
Just a bit different approach:
$$fracmathrmdymathrmdx=frac1y-3$$
Let $u=y-3$, then we have that
$$fracmathrmdumathrmdx=fracmathrmdymathrmdx$$
So
$$fracmathrmdumathrmdx=frac1u$$
$$ufracmathrmdumathrmdx=1$$
$$2ufracmathrmdumathrmdx=2$$
$$fracmathrmdu^2mathrmdx=2$$
$$u^2=2x+C$$
And if we use the initial conditions now:
$$y(2)=u(2)+3=5$$
$$u(2)=2$$
So
$$4=4+C$$
$$C=0$$
Back to the equation:
$$u^2=2x$$
$$u=sqrt2x$$
$$y-3=sqrt2x$$
$$y(x)=sqrt2x+3$$
answered 2 days ago
Botond
3,8702532
add a comment |Â
up vote
0
down vote
Just a bit different approach:
$$fracmathrmdymathrmdx=frac1y-3$$
Let $u=y-3$, then we have that
$$fracmathrmdumathrmdx=fracmathrmdymathrmdx$$
So
$$fracmathrmdumathrmdx=frac1u$$
$$ufracmathrmdumathrmdx=1$$
$$2ufracmathrmdumathrmdx=2$$
$$fracmathrmdu^2mathrmdx=2$$
$$u^2=2x+C$$
And if we use the initial conditions now:
$$y(2)=u(2)+3=5$$
$$u(2)=2$$
So
$$4=4+C$$
$$C=0$$
Back to the equation:
$$u^2=2x$$
$$u=sqrt2x$$
$$y-3=sqrt2x$$
$$y(x)=sqrt2x+3$$
answered 2 days ago
Botond
3,8702532
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Just a bit different approach:
$$fracmathrmdymathrmdx=frac1y-3$$
Let $u=y-3$, then we have that
$$fracmathrmdumathrmdx=fracmathrmdymathrmdx$$
So
$$fracmathrmdumathrmdx=frac1u$$
$$ufracmathrmdumathrmdx=1$$
$$2ufracmathrmdumathrmdx=2$$
$$fracmathrmdu^2mathrmdx=2$$
$$u^2=2x+C$$
And if we use the initial conditions now:
$$y(2)=u(2)+3=5$$
$$u(2)=2$$
So
$$4=4+C$$
$$C=0$$
Back to the equation:
$$u^2=2x$$
$$u=sqrt2x$$
$$y-3=sqrt2x$$
$$y(x)=sqrt2x+3$$
answered 2 days ago
Botond
3,8702532
Just a bit different approach:
$$fracmathrmdymathrmdx=frac1y-3$$
Let $u=y-3$, then we have that
$$fracmathrmdumathrmdx=fracmathrmdymathrmdx$$
So
$$fracmathrmdumathrmdx=frac1u$$
$$ufracmathrmdumathrmdx=1$$
$$2ufracmathrmdumathrmdx=2$$
$$fracmathrmdu^2mathrmdx=2$$
$$u^2=2x+C$$
And if we use the initial conditions now:
$$y(2)=u(2)+3=5$$
$$u(2)=2$$
So
$$4=4+C$$
$$C=0$$
Back to the equation:
$$u^2=2x$$
$$u=sqrt2x$$
$$y-3=sqrt2x$$
$$y(x)=sqrt2x+3$$
answered 2 days ago
Botond
3,8702532
answered 2 days ago
Botond
3,8702532
answered 2 days ago
Botond
3,8702532
answered 2 days ago
Botond
3,8702532
3,8702532
add a comment |Â
add a comment |Â
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Sometimes it is just not possible or worth the effort to get an explicit solution in terms of $y$
– Gobabis
2 days ago
What quadratic equation have you got? After seperating the variables the equation is $(y-3) dy=dxRightarrow int (y-3) dy=int dx$
– callculus
2 days ago
Have you solved the problem by using the hint? If not, give a reply. If yes, give a reply as well.
– callculus
2 days ago
@Gobabis It doesn´t take much effort to solve the equation.
– callculus
2 days ago