Mixing
Solve the differential equation $frac{dy}{dx}=5+xy+2x+2y$
$begingroup$
Solve the differential equation $$frac{dy}{dx}=5+xy+2x+2y$$ Given $y(0)=0$
My try:
The given equation can be written as:
$$frac{dy}{dx}=1+(x+2)(y+2)$$
Letting $X=x+2$ and $Y=y+2$ we get $dy=dY$ and $dx=dX$
So the equation becomes
$$frac{dY}{dX}=1+XY$$ or
$$frac{dY}{dX}-XY=1$$ which is a linear differential equation of first order:
The integrating factor is
$$I(X)=e^{frac{-X^2}{2}}$$
The solution is:
$$Ye^{frac{-X^2}{2}}=int e^{frac{-X^2}{2}}dx+C$$
How to continue here?
algebra-precalculus ordinary-differential-equations gaussian-integral
asked Jan 3 at 4:51
Umesh shankarUmesh shankar
2,61431219
$endgroup$
add a comment |
$begingroup$
Solve the differential equation $$frac{dy}{dx}=5+xy+2x+2y$$ Given $y(0)=0$
My try:
The given equation can be written as:
$$frac{dy}{dx}=1+(x+2)(y+2)$$
Letting $X=x+2$ and $Y=y+2$ we get $dy=dY$ and $dx=dX$
So the equation becomes
$$frac{dY}{dX}=1+XY$$ or
$$frac{dY}{dX}-XY=1$$ which is a linear differential equation of first order:
The integrating factor is
$$I(X)=e^{frac{-X^2}{2}}$$
The solution is:
$$Ye^{frac{-X^2}{2}}=int e^{frac{-X^2}{2}}dx+C$$
How to continue here?
algebra-precalculus ordinary-differential-equations gaussian-integral
asked Jan 3 at 4:51
Umesh shankarUmesh shankar
2,61431219
$endgroup$
add a comment |
$begingroup$
Solve the differential equation $$frac{dy}{dx}=5+xy+2x+2y$$ Given $y(0)=0$
My try:
The given equation can be written as:
$$frac{dy}{dx}=1+(x+2)(y+2)$$
Letting $X=x+2$ and $Y=y+2$ we get $dy=dY$ and $dx=dX$
So the equation becomes
$$frac{dY}{dX}=1+XY$$ or
$$frac{dY}{dX}-XY=1$$ which is a linear differential equation of first order:
The integrating factor is
$$I(X)=e^{frac{-X^2}{2}}$$
The solution is:
$$Ye^{frac{-X^2}{2}}=int e^{frac{-X^2}{2}}dx+C$$
How to continue here?
algebra-precalculus ordinary-differential-equations gaussian-integral
asked Jan 3 at 4:51
Umesh shankarUmesh shankar
2,61431219
$endgroup$
Solve the differential equation $$frac{dy}{dx}=5+xy+2x+2y$$ Given $y(0)=0$
My try:
The given equation can be written as:
$$frac{dy}{dx}=1+(x+2)(y+2)$$
Letting $X=x+2$ and $Y=y+2$ we get $dy=dY$ and $dx=dX$
So the equation becomes
$$frac{dY}{dX}=1+XY$$ or
$$frac{dY}{dX}-XY=1$$ which is a linear differential equation of first order:
The integrating factor is
$$I(X)=e^{frac{-X^2}{2}}$$
The solution is:
$$Ye^{frac{-X^2}{2}}=int e^{frac{-X^2}{2}}dx+C$$
How to continue here?
algebra-precalculus ordinary-differential-equations gaussian-integral
algebra-precalculus ordinary-differential-equations gaussian-integral
asked Jan 3 at 4:51
Umesh shankarUmesh shankar
2,61431219
asked Jan 3 at 4:51
Umesh shankarUmesh shankar
2,61431219
asked Jan 3 at 4:51
Umesh shankarUmesh shankar
2,61431219
asked Jan 3 at 4:51
Umesh shankarUmesh shankar
2,61431219
asked Jan 3 at 4:51
Umesh shankarUmesh shankar
2,61431219
2,61431219
add a comment |
add a comment |
2 Answers
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oldest
votes
$begingroup$
The integral you have on the RHS is called the error function, defined by $$text{erf}(z)=frac2{sqrtpi}int_0^z e^{-x^2}mathrm dx$$So in this case, you have found $sqrtfrac{pi}2text{erf}(X/sqrt2)$. Therefore the solution in terms of $X,Y$ is $$Y=sqrtfracpi2e^{frac{X^2}2}text{erf}left(frac X{sqrt2}right)+Ce^{frac{X^2}2}$$Then sub back $x$ and $y$ to get the final solution.
answered Jan 3 at 5:07
John DoeJohn Doe
11.1k11238
$endgroup$
add a comment |
$begingroup$
Divide through by $e^{-X^2/2}$ and back substitute.
answered Jan 3 at 5:07
Guacho PerezGuacho Perez
3,91411132
$endgroup$
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
$begingroup$
The integral you have on the RHS is called the error function, defined by $$text{erf}(z)=frac2{sqrtpi}int_0^z e^{-x^2}mathrm dx$$So in this case, you have found $sqrtfrac{pi}2text{erf}(X/sqrt2)$. Therefore the solution in terms of $X,Y$ is $$Y=sqrtfracpi2e^{frac{X^2}2}text{erf}left(frac X{sqrt2}right)+Ce^{frac{X^2}2}$$Then sub back $x$ and $y$ to get the final solution.
answered Jan 3 at 5:07
John DoeJohn Doe
11.1k11238
$endgroup$
add a comment |
$begingroup$
The integral you have on the RHS is called the error function, defined by $$text{erf}(z)=frac2{sqrtpi}int_0^z e^{-x^2}mathrm dx$$So in this case, you have found $sqrtfrac{pi}2text{erf}(X/sqrt2)$. Therefore the solution in terms of $X,Y$ is $$Y=sqrtfracpi2e^{frac{X^2}2}text{erf}left(frac X{sqrt2}right)+Ce^{frac{X^2}2}$$Then sub back $x$ and $y$ to get the final solution.
answered Jan 3 at 5:07
John DoeJohn Doe
11.1k11238
$endgroup$
add a comment |
$begingroup$
The integral you have on the RHS is called the error function, defined by $$text{erf}(z)=frac2{sqrtpi}int_0^z e^{-x^2}mathrm dx$$So in this case, you have found $sqrtfrac{pi}2text{erf}(X/sqrt2)$. Therefore the solution in terms of $X,Y$ is $$Y=sqrtfracpi2e^{frac{X^2}2}text{erf}left(frac X{sqrt2}right)+Ce^{frac{X^2}2}$$Then sub back $x$ and $y$ to get the final solution.
answered Jan 3 at 5:07
John DoeJohn Doe
11.1k11238
$endgroup$
The integral you have on the RHS is called the error function, defined by $$text{erf}(z)=frac2{sqrtpi}int_0^z e^{-x^2}mathrm dx$$So in this case, you have found $sqrtfrac{pi}2text{erf}(X/sqrt2)$. Therefore the solution in terms of $X,Y$ is $$Y=sqrtfracpi2e^{frac{X^2}2}text{erf}left(frac X{sqrt2}right)+Ce^{frac{X^2}2}$$Then sub back $x$ and $y$ to get the final solution.
answered Jan 3 at 5:07
John DoeJohn Doe
11.1k11238
answered Jan 3 at 5:07
John DoeJohn Doe
11.1k11238
answered Jan 3 at 5:07
John DoeJohn Doe
11.1k11238
answered Jan 3 at 5:07
John DoeJohn Doe
11.1k11238
11.1k11238
add a comment |
add a comment |
$begingroup$
Divide through by $e^{-X^2/2}$ and back substitute.
answered Jan 3 at 5:07
Guacho PerezGuacho Perez
3,91411132
$endgroup$
add a comment |
$begingroup$
Divide through by $e^{-X^2/2}$ and back substitute.
answered Jan 3 at 5:07
Guacho PerezGuacho Perez
3,91411132
$endgroup$
add a comment |
$begingroup$
Divide through by $e^{-X^2/2}$ and back substitute.
answered Jan 3 at 5:07
Guacho PerezGuacho Perez
3,91411132
$endgroup$
Divide through by $e^{-X^2/2}$ and back substitute.
answered Jan 3 at 5:07
Guacho PerezGuacho Perez
3,91411132
answered Jan 3 at 5:07
Guacho PerezGuacho Perez
3,91411132
answered Jan 3 at 5:07
Guacho PerezGuacho Perez
3,91411132
answered Jan 3 at 5:07
Guacho PerezGuacho Perez
3,91411132
3,91411132
add a comment |
add a comment |
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