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Value of $f(0)$ in differential equation.
If $f(pi) = 2$ and $displaystyle int^{pi}_{0}bigg[f(x)+f''(x)bigg]sin(x)dx=5.$
Then value of $f(0).$
Given that function $f(x)$ is continuous in $[0,pi]$
Try: $$int bigg[f(x)+f''(x)bigg]sin xdx$$
$$ = int f(x)sin xdx+sin xf'(x)+int cos xf'(x)dx$$
$$ = int f(x)sin xdx+sin xf'(x)+cos xf(x)-int sin x f(x)dx$$
So $$int^{pi}_{0}bigg[f(x)+f''(x)bigg]sin(x)dx = sin xf'(x)+cos x f(x) = 5.$$
So $$sin xf'(x)+cos xf(x) = 5$$
Could some help me how i solve above differential equation. Thanks
differential-equations differential
asked Nov 22 '18 at 9:44
D TiwariD Tiwari
5,3802630
add a comment |
If $f(pi) = 2$ and $displaystyle int^{pi}_{0}bigg[f(x)+f''(x)bigg]sin(x)dx=5.$
Then value of $f(0).$
Given that function $f(x)$ is continuous in $[0,pi]$
Try: $$int bigg[f(x)+f''(x)bigg]sin xdx$$
$$ = int f(x)sin xdx+sin xf'(x)+int cos xf'(x)dx$$
$$ = int f(x)sin xdx+sin xf'(x)+cos xf(x)-int sin x f(x)dx$$
So $$int^{pi}_{0}bigg[f(x)+f''(x)bigg]sin(x)dx = sin xf'(x)+cos x f(x) = 5.$$
So $$sin xf'(x)+cos xf(x) = 5$$
Could some help me how i solve above differential equation. Thanks
differential-equations differential
asked Nov 22 '18 at 9:44
D TiwariD Tiwari
5,3802630
1
You should have done with the limits.
– Yadati Kiran
Nov 22 '18 at 9:56
add a comment |
If $f(pi) = 2$ and $displaystyle int^{pi}_{0}bigg[f(x)+f''(x)bigg]sin(x)dx=5.$
Then value of $f(0).$
Given that function $f(x)$ is continuous in $[0,pi]$
Try: $$int bigg[f(x)+f''(x)bigg]sin xdx$$
$$ = int f(x)sin xdx+sin xf'(x)+int cos xf'(x)dx$$
$$ = int f(x)sin xdx+sin xf'(x)+cos xf(x)-int sin x f(x)dx$$
So $$int^{pi}_{0}bigg[f(x)+f''(x)bigg]sin(x)dx = sin xf'(x)+cos x f(x) = 5.$$
So $$sin xf'(x)+cos xf(x) = 5$$
Could some help me how i solve above differential equation. Thanks
differential-equations differential
asked Nov 22 '18 at 9:44
D TiwariD Tiwari
5,3802630
If $f(pi) = 2$ and $displaystyle int^{pi}_{0}bigg[f(x)+f''(x)bigg]sin(x)dx=5.$
Then value of $f(0).$
Given that function $f(x)$ is continuous in $[0,pi]$
Try: $$int bigg[f(x)+f''(x)bigg]sin xdx$$
$$ = int f(x)sin xdx+sin xf'(x)+int cos xf'(x)dx$$
$$ = int f(x)sin xdx+sin xf'(x)+cos xf(x)-int sin x f(x)dx$$
So $$int^{pi}_{0}bigg[f(x)+f''(x)bigg]sin(x)dx = sin xf'(x)+cos x f(x) = 5.$$
So $$sin xf'(x)+cos xf(x) = 5$$
Could some help me how i solve above differential equation. Thanks
differential-equations differential
differential-equations differential
asked Nov 22 '18 at 9:44
D TiwariD Tiwari
5,3802630
asked Nov 22 '18 at 9:44
D TiwariD Tiwari
5,3802630
asked Nov 22 '18 at 9:44
D TiwariD Tiwari
5,3802630
asked Nov 22 '18 at 9:44
D TiwariD Tiwari
5,3802630
asked Nov 22 '18 at 9:44
D TiwariD Tiwari
5,3802630
5,3802630
1
You should have done with the limits.
– Yadati Kiran
Nov 22 '18 at 9:56
add a comment |
1
You should have done with the limits.
– Yadati Kiran
Nov 22 '18 at 9:56
1
1
You should have done with the limits.
– Yadati Kiran
Nov 22 '18 at 9:56
You should have done with the limits.
– Yadati Kiran
Nov 22 '18 at 9:56
add a comment |
2 Answers
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Integrate by part twice. $int_0^{pi} f''(x)sin, x , dx= -int_0^{pi} f'(x)cos, x , dx=f(0) -f(pi)(-1)-int_0^{pi} f(x)sin, x , dx$ so we get $5=f(0) +f(pi)$. Hence $f(0)=3$.
answered Nov 22 '18 at 9:51
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
add a comment |
In fact, you have obtained that $$f'(x)sin x+f(x)cos x|_{0}^{pi}=5$$which means that $$-f(pi)+f(0)=5$$from which we obtain$$f(0)=7$$
answered Nov 22 '18 at 10:16
Mostafa AyazMostafa Ayaz
14.5k3937
add a comment |
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2 Answers
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2 Answers
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Integrate by part twice. $int_0^{pi} f''(x)sin, x , dx= -int_0^{pi} f'(x)cos, x , dx=f(0) -f(pi)(-1)-int_0^{pi} f(x)sin, x , dx$ so we get $5=f(0) +f(pi)$. Hence $f(0)=3$.
answered Nov 22 '18 at 9:51
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
add a comment |
Integrate by part twice. $int_0^{pi} f''(x)sin, x , dx= -int_0^{pi} f'(x)cos, x , dx=f(0) -f(pi)(-1)-int_0^{pi} f(x)sin, x , dx$ so we get $5=f(0) +f(pi)$. Hence $f(0)=3$.
answered Nov 22 '18 at 9:51
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
add a comment |
Integrate by part twice. $int_0^{pi} f''(x)sin, x , dx= -int_0^{pi} f'(x)cos, x , dx=f(0) -f(pi)(-1)-int_0^{pi} f(x)sin, x , dx$ so we get $5=f(0) +f(pi)$. Hence $f(0)=3$.
answered Nov 22 '18 at 9:51
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
Integrate by part twice. $int_0^{pi} f''(x)sin, x , dx= -int_0^{pi} f'(x)cos, x , dx=f(0) -f(pi)(-1)-int_0^{pi} f(x)sin, x , dx$ so we get $5=f(0) +f(pi)$. Hence $f(0)=3$.
answered Nov 22 '18 at 9:51
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
answered Nov 22 '18 at 9:51
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
answered Nov 22 '18 at 9:51
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
answered Nov 22 '18 at 9:51
Kavi Rama MurthyKavi Rama Murthy
52.1k32055
52.1k32055
add a comment |
add a comment |
In fact, you have obtained that $$f'(x)sin x+f(x)cos x|_{0}^{pi}=5$$which means that $$-f(pi)+f(0)=5$$from which we obtain$$f(0)=7$$
answered Nov 22 '18 at 10:16
Mostafa AyazMostafa Ayaz
14.5k3937
add a comment |
In fact, you have obtained that $$f'(x)sin x+f(x)cos x|_{0}^{pi}=5$$which means that $$-f(pi)+f(0)=5$$from which we obtain$$f(0)=7$$
answered Nov 22 '18 at 10:16
Mostafa AyazMostafa Ayaz
14.5k3937
add a comment |
In fact, you have obtained that $$f'(x)sin x+f(x)cos x|_{0}^{pi}=5$$which means that $$-f(pi)+f(0)=5$$from which we obtain$$f(0)=7$$
answered Nov 22 '18 at 10:16
Mostafa AyazMostafa Ayaz
14.5k3937
In fact, you have obtained that $$f'(x)sin x+f(x)cos x|_{0}^{pi}=5$$which means that $$-f(pi)+f(0)=5$$from which we obtain$$f(0)=7$$
answered Nov 22 '18 at 10:16
Mostafa AyazMostafa Ayaz
14.5k3937
answered Nov 22 '18 at 10:16
Mostafa AyazMostafa Ayaz
14.5k3937
answered Nov 22 '18 at 10:16
Mostafa AyazMostafa Ayaz
14.5k3937
answered Nov 22 '18 at 10:16
Mostafa AyazMostafa Ayaz
14.5k3937
14.5k3937
add a comment |
add a comment |
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You should have done with the limits.
– Yadati Kiran
Nov 22 '18 at 9:56