Mixing
Differential Equations Was abe to show $y'=-4x$ $y''=-4$ and when substitute $=0$ [closed]
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$(1-x^2)y''-xy'+4y=0$ Show that $1-2x^2$ is a solution try to find another using Wronskian method.
Was abe to show $y'=-4x$ $y''=-4$ and when substitute $=0$
differential-equations
asked Jul 22 at 12:57
Liz
396
closed as off-topic by Brian Borchers, Claude Leibovici, Tyrone, amWhy, Adrian Keister Jul 24 at 13:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brian Borchers, Claude Leibovici, Tyrone, amWhy
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up vote
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favorite
$(1-x^2)y''-xy'+4y=0$ Show that $1-2x^2$ is a solution try to find another using Wronskian method.
Was abe to show $y'=-4x$ $y''=-4$ and when substitute $=0$
differential-equations
asked Jul 22 at 12:57
Liz
396
closed as off-topic by Brian Borchers, Claude Leibovici, Tyrone, amWhy, Adrian Keister Jul 24 at 13:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brian Borchers, Claude Leibovici, Tyrone, amWhy
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$(1-x^2)y''-xy'+4y=0$ Show that $1-2x^2$ is a solution try to find another using Wronskian method.
Was abe to show $y'=-4x$ $y''=-4$ and when substitute $=0$
differential-equations
asked Jul 22 at 12:57
Liz
396
$(1-x^2)y''-xy'+4y=0$ Show that $1-2x^2$ is a solution try to find another using Wronskian method.
Was abe to show $y'=-4x$ $y''=-4$ and when substitute $=0$
differential-equations
asked Jul 22 at 12:57
Liz
396
edited Jul 22 at 20:57
asked Jul 22 at 12:57
Liz
396
asked Jul 22 at 12:57
Liz
396
asked Jul 22 at 12:57
Liz
396
396
closed as off-topic by Brian Borchers, Claude Leibovici, Tyrone, amWhy, Adrian Keister Jul 24 at 13:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brian Borchers, Claude Leibovici, Tyrone, amWhy
closed as off-topic by Brian Borchers, Claude Leibovici, Tyrone, amWhy, Adrian Keister Jul 24 at 13:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brian Borchers, Claude Leibovici, Tyrone, amWhy
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2 Answers
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Hint
There are two ways to calculate the Wronskian
$$beginalign
&(1)~W(y_1,y_2)(t)~=~beginvmatrixy_1(t) & y_2(t)\y_1'(t) & y_2'(t)endvmatrix~=~y_1(t)y_2'(t)-y_2(t)y_1'(t)\
&(2)~W(y_1,y_2)(t)~=~e^-int p(x)mathrmdx
endalign$$
You already know one solution, $y_1$, of the differential equation.
answered Jul 22 at 13:24
mrtaurho
740219
1
What is $p$ in your case? I think you should include it.
– Botond
Jul 22 at 13:30
1
The coefficient of the first derivative in the normal form, when I remember it right. In this case $p(x)=-fracx1-x^2$.
– mrtaurho
Jul 22 at 13:44
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up vote
2
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Hint: With $p(x)=dfrac-x1-x^2$ we write
$$W(x)=expleft(-intdfrac-x1-x^2right)dx=dfrac1sqrt1-x^2$$
for $|x|<1$. Then solve the integral
$$y_2=(1-2x^2)intdfracW(x)(1-2x^2)^2dx$$
answered Jul 22 at 16:10
Nosrati
19.4k41544
does the negative flip the equation. 1/sqrt(x^2-1) ???
– Liz
Jul 23 at 13:21
I don't understand your mean. With $|x|<1$ then $W(x)=dfrac1sqrt1-x^2$ and for $|x|>1$ then $W(x)=dfrac1sqrtx^2-1$.
– Nosrati
Jul 23 at 13:24
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Hint
There are two ways to calculate the Wronskian
$$beginalign
&(1)~W(y_1,y_2)(t)~=~beginvmatrixy_1(t) & y_2(t)\y_1'(t) & y_2'(t)endvmatrix~=~y_1(t)y_2'(t)-y_2(t)y_1'(t)\
&(2)~W(y_1,y_2)(t)~=~e^-int p(x)mathrmdx
endalign$$
You already know one solution, $y_1$, of the differential equation.
answered Jul 22 at 13:24
mrtaurho
740219
1
What is $p$ in your case? I think you should include it.
– Botond
Jul 22 at 13:30
1
The coefficient of the first derivative in the normal form, when I remember it right. In this case $p(x)=-fracx1-x^2$.
– mrtaurho
Jul 22 at 13:44
add a comment |Â
up vote
2
down vote
Hint
There are two ways to calculate the Wronskian
$$beginalign
&(1)~W(y_1,y_2)(t)~=~beginvmatrixy_1(t) & y_2(t)\y_1'(t) & y_2'(t)endvmatrix~=~y_1(t)y_2'(t)-y_2(t)y_1'(t)\
&(2)~W(y_1,y_2)(t)~=~e^-int p(x)mathrmdx
endalign$$
You already know one solution, $y_1$, of the differential equation.
answered Jul 22 at 13:24
mrtaurho
740219
1
What is $p$ in your case? I think you should include it.
– Botond
Jul 22 at 13:30
1
The coefficient of the first derivative in the normal form, when I remember it right. In this case $p(x)=-fracx1-x^2$.
– mrtaurho
Jul 22 at 13:44
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint
There are two ways to calculate the Wronskian
$$beginalign
&(1)~W(y_1,y_2)(t)~=~beginvmatrixy_1(t) & y_2(t)\y_1'(t) & y_2'(t)endvmatrix~=~y_1(t)y_2'(t)-y_2(t)y_1'(t)\
&(2)~W(y_1,y_2)(t)~=~e^-int p(x)mathrmdx
endalign$$
You already know one solution, $y_1$, of the differential equation.
answered Jul 22 at 13:24
mrtaurho
740219
Hint
There are two ways to calculate the Wronskian
$$beginalign
&(1)~W(y_1,y_2)(t)~=~beginvmatrixy_1(t) & y_2(t)\y_1'(t) & y_2'(t)endvmatrix~=~y_1(t)y_2'(t)-y_2(t)y_1'(t)\
&(2)~W(y_1,y_2)(t)~=~e^-int p(x)mathrmdx
endalign$$
You already know one solution, $y_1$, of the differential equation.
answered Jul 22 at 13:24
mrtaurho
740219
answered Jul 22 at 13:24
mrtaurho
740219
answered Jul 22 at 13:24
mrtaurho
740219
answered Jul 22 at 13:24
mrtaurho
740219
740219
1
What is $p$ in your case? I think you should include it.
– Botond
Jul 22 at 13:30
1
The coefficient of the first derivative in the normal form, when I remember it right. In this case $p(x)=-fracx1-x^2$.
– mrtaurho
Jul 22 at 13:44
add a comment |Â
1
What is $p$ in your case? I think you should include it.
– Botond
Jul 22 at 13:30
1
The coefficient of the first derivative in the normal form, when I remember it right. In this case $p(x)=-fracx1-x^2$.
– mrtaurho
Jul 22 at 13:44
1
1
What is $p$ in your case? I think you should include it.
– Botond
Jul 22 at 13:30
What is $p$ in your case? I think you should include it.
– Botond
Jul 22 at 13:30
1
1
The coefficient of the first derivative in the normal form, when I remember it right. In this case $p(x)=-fracx1-x^2$.
– mrtaurho
Jul 22 at 13:44
The coefficient of the first derivative in the normal form, when I remember it right. In this case $p(x)=-fracx1-x^2$.
– mrtaurho
Jul 22 at 13:44
add a comment |Â
up vote
2
down vote
Hint: With $p(x)=dfrac-x1-x^2$ we write
$$W(x)=expleft(-intdfrac-x1-x^2right)dx=dfrac1sqrt1-x^2$$
for $|x|<1$. Then solve the integral
$$y_2=(1-2x^2)intdfracW(x)(1-2x^2)^2dx$$
answered Jul 22 at 16:10
Nosrati
19.4k41544
does the negative flip the equation. 1/sqrt(x^2-1) ???
– Liz
Jul 23 at 13:21
I don't understand your mean. With $|x|<1$ then $W(x)=dfrac1sqrt1-x^2$ and for $|x|>1$ then $W(x)=dfrac1sqrtx^2-1$.
– Nosrati
Jul 23 at 13:24
add a comment |Â
up vote
2
down vote
Hint: With $p(x)=dfrac-x1-x^2$ we write
$$W(x)=expleft(-intdfrac-x1-x^2right)dx=dfrac1sqrt1-x^2$$
for $|x|<1$. Then solve the integral
$$y_2=(1-2x^2)intdfracW(x)(1-2x^2)^2dx$$
answered Jul 22 at 16:10
Nosrati
19.4k41544
does the negative flip the equation. 1/sqrt(x^2-1) ???
– Liz
Jul 23 at 13:21
I don't understand your mean. With $|x|<1$ then $W(x)=dfrac1sqrt1-x^2$ and for $|x|>1$ then $W(x)=dfrac1sqrtx^2-1$.
– Nosrati
Jul 23 at 13:24
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint: With $p(x)=dfrac-x1-x^2$ we write
$$W(x)=expleft(-intdfrac-x1-x^2right)dx=dfrac1sqrt1-x^2$$
for $|x|<1$. Then solve the integral
$$y_2=(1-2x^2)intdfracW(x)(1-2x^2)^2dx$$
answered Jul 22 at 16:10
Nosrati
19.4k41544
Hint: With $p(x)=dfrac-x1-x^2$ we write
$$W(x)=expleft(-intdfrac-x1-x^2right)dx=dfrac1sqrt1-x^2$$
for $|x|<1$. Then solve the integral
$$y_2=(1-2x^2)intdfracW(x)(1-2x^2)^2dx$$
answered Jul 22 at 16:10
Nosrati
19.4k41544
answered Jul 22 at 16:10
Nosrati
19.4k41544
answered Jul 22 at 16:10
Nosrati
19.4k41544
answered Jul 22 at 16:10
Nosrati
19.4k41544
19.4k41544
does the negative flip the equation. 1/sqrt(x^2-1) ???
– Liz
Jul 23 at 13:21
I don't understand your mean. With $|x|<1$ then $W(x)=dfrac1sqrt1-x^2$ and for $|x|>1$ then $W(x)=dfrac1sqrtx^2-1$.
– Nosrati
Jul 23 at 13:24
add a comment |Â
does the negative flip the equation. 1/sqrt(x^2-1) ???
– Liz
Jul 23 at 13:21
I don't understand your mean. With $|x|<1$ then $W(x)=dfrac1sqrt1-x^2$ and for $|x|>1$ then $W(x)=dfrac1sqrtx^2-1$.
– Nosrati
Jul 23 at 13:24
does the negative flip the equation. 1/sqrt(x^2-1) ???
– Liz
Jul 23 at 13:21
does the negative flip the equation. 1/sqrt(x^2-1) ???
– Liz
Jul 23 at 13:21
I don't understand your mean. With $|x|<1$ then $W(x)=dfrac1sqrt1-x^2$ and for $|x|>1$ then $W(x)=dfrac1sqrtx^2-1$.
– Nosrati
Jul 23 at 13:24
I don't understand your mean. With $|x|<1$ then $W(x)=dfrac1sqrt1-x^2$ and for $|x|>1$ then $W(x)=dfrac1sqrtx^2-1$.
– Nosrati
Jul 23 at 13:24
add a comment |Â