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Exercise 9.5 - Chapter 9 Differential Equations class 12 ncert solutions Maths - SaraNextGen [2024-2025]


Updated By SaraNextGen
On April 24, 2024, 11:35 AM

Question 1:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7857/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m501d51d2.gif

Answer:

The given differential equation i.e., (x2 + xydy = (x2 + y2dx can be written as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7857/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_468af196.gif

This shows that equation (1) is a homogeneous equation.

To solve it, we make the substitution as:

vx

Differentiating both sides with respect to x, we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7857/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_224e91db.gif

Substituting the values of v and https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7857/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m72221781.gif in equation (1), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7857/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_61b044ec.gif

Integrating both sides, we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7857/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_3cb673b1.gif

This is the required solution of the given differential equation.

Question 2:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7859/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m973c521.gif

Answer:

The given differential equation is:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7859/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m670d98c.gif

Thus, the given equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

Differentiating both sides with respect to x, we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7859/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_224e91db.gif

Substituting the values of y and https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7859/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m72221781.gif in equation (1), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7859/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_3a99fd6d.gif

Integrating both sides, we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7859/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m33b8352d.gif

This is the required solution of the given differential equation.

Question 3:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7861/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_37f98764.gif

Answer:

The given differential equation is:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7861/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_37f98764.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7861/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_26b48392.gif

Thus, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7861/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m6305e560.gif

Substituting the values of y and https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7861/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m72221781.gif in equation (1), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7861/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_388a264a.gif

Integrating both sides, we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7861/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_20d6a717.gif

This is the required solution of the given differential equation.

Question 4:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7863/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_48d97c5a.gif

Answer:

The given differential equation is:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7863/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_48d97c5a.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7863/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m1ef08d50.gif

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7863/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m6305e560.gif

Substituting the values of y and https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7863/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m72221781.gif in equation (1), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7863/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_52ac7312.gif

Integrating both sides, we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7863/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_5d1f8e64.gif

This is the required solution of the given differential equation.

Question 5:

https://img-nm.mnimgs.com/img/study_content/editlive_ncert/33/2012_02_21_18_18_00/mathmlequation2229679458588616949.png

Answer:

The given differential equation is:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7865/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_me1b4e05.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7865/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m546757fb.gif

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7865/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m29d2fdbc.gif

Substituting the values of y and https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7865/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m72221781.gif in equation (1), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7865/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_73686725.gif

Integrating both sides, we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7865/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_66457b63.gif

This is the required solution for the given differential equation.

Question 6:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7867/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_6a7383fe.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/content_ck_images/images/Correction.gif

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7867/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m6305e560.gif

Substituting the values of v and https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7867/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m72221781.gif in equation (1), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7867/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m33584c21.gif

Integrating both sides, we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7867/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_6201ef90.gif

This is the required solution of the given differential equation.

Question 7:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7871/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_6fd1a0e.gif

Answer:

The given differential equation is:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7871/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m69384e9f.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7871/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_7ea34bd0.gif

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7871/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m52d3f8f9.gif

Substituting the values of y and https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7871/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m72221781.gif in equation (1), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7871/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_6c6223.gif

Integrating both sides, we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7871/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_1b66b121.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7871/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m7c82eb1c.gif

This is the required solution of the given differential equation.

Question 8:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7874/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m2fbda755.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7874/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m31e95b94.gif

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7874/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m6305e560.gif

Substituting the values of y and https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7874/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m72221781.gif in equation (1), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7874/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_7061f891.gif

Integrating both sides, we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7874/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_5f897286.gif

This is the required solution of the given differential equation.

Question 9:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7877/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_3e600719.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7877/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_4a386dc4.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7877/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_71d2abdd.gif

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7877/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m6ca14e08.gif

Substituting the values of y and https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7877/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m72221781.gif in equation (1), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7877/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m50de6cb2.gif

Integrating both sides, we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7877/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_mc87225f.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7877/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m33f40725.gif

Therefore, equation (1) becomes:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7877/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m212ae962.gif

This is the required solution of the given differential equation.

Question 10:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7880/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_1684f6df.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7880/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m45bf726b.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7880/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_18adc76b.gif

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vy

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7880/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_afe66dd.gif

Substituting the values of x and https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7880/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_7bd1cf62.gif in equation (1), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7880/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_6e6cbffd.gif

Integrating both sides, we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7880/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_35c908e9.gif

This is the required solution of the given differential equation.

Question 11:

https://img-nm.mnimgs.com/img/study_content/editlive_ncert/33/2012_02_21_18_20_53/mathmlequation6844457755488916695.png

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7881/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_65a02c22.gif

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7881/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m6305e560.gif

Substituting the values of y and https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7881/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m72221781.gif in equation (1), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7881/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_6d73669a.gif

Integrating both sides, we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7881/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m7b817b7d.gif

Now, y = 1 at x = 1.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7881/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_d018409.gif

Substituting the value of 2k in equation (2), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7881/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_7629563.gif

This is the required solution of the given differential equation.

Question 12:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7883/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m2c45a64.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7883/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_379506c9.gif

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7883/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m6305e560.gif

Substituting the values of y and https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7883/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m72221781.gif in equation (1), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7883/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m2b742fb5.gif

Integrating both sides, we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7883/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m4ddd2dd7.gif

Now, y = 1 at x = 1.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7883/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_330f2f01.gif

Substituting https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7883/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_261def3d.gif in equation (2), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7883/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_351c847a.gif

This is the required solution of the given differential equation.

Question 13:

https://img-nm.mnimgs.com/img/study_content/editlive_ncert/33/2012_02_21_18_33_42/mathmlequation4401542383907660145.png

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7885/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_11c56719.gif

Therefore, the given differential equation is a homogeneous equation.

To solve this differential equation, we make the substitution as:

vx

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7885/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_3dff1adc.gif

Substituting the values of y and https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7885/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m72221781.gif in equation (1), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7885/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m5578fce1.gif

Integrating both sides, we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7885/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m7eda5bd.gif

Now, https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7885/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m442f6de2.gif .

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7885/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_mfe44a1f.gif

Substituting C = e in equation (2), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7885/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_56a0dfd5.gif

This is the required solution of the given differential equation.

Question 14:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7887/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_2a64122f.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7887/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_47f61fb1.gif

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7887/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m6305e560.gif

Substituting the values of y and https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7887/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m72221781.gif in equation (1), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7887/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m1fb6564d.gif

Integrating both sides, we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7887/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m749f6ef3.gif

This is the required solution of the given differential equation.

Now, y = 0 at x = 1.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7887/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m2439f1a8.gif

Substituting C = e in equation (2), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7887/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_123984f5.gif

This is the required solution of the given differential equation.

Question 15:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7888/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m6d8af437.gif

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7888/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_1d36072b.gif

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

vx

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7888/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m6305e560.gif

Substituting the value of y and https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7888/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m72221781.gif in equation (1), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7888/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_708e8096.gif

Integrating both sides, we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7888/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_598c85d8.gif

Now, y = 2 at x = 1.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7888/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_445b41c7.gif

Substituting C = –1 in equation (2), we get:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7888/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m3904717f.gif

This is the required solution of the given differential equation.

Question 16:

A homogeneous differential equation of the form https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7890/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_397d3c70.gif can be solved by making the substitution

A. y = vx

B. v = yx

C. vy

D. x = v

Answer:

For solving the homogeneous equation of the formhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7890/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_397d3c70.gif , we need to make the substitution as x = vy.

Hence, the correct Answer is C.

Question 17:

Which of the following is a homogeneous differential equation?

A. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7892/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_4d154b48.gif

B. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7892/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_3f584b72.gif

C. https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7892/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_65846662.gif

D. https://img-nm.mnimgs.com/img/study_content/editlive_ncert/33/2012_02_21_18_36_00/mathmlequation7821902014782140484.png

Answer:

Function F(xy) is said to be the homogenous function of degree n, if

F(λx, λy) = λn F(xy) for any non-zero constant (λ).

Consider the equation given in alternativeD:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/238/7892/NCERT_12-11-08_Gopal_12_Maths_Ex-9.1_12_MNK_SG_html_m40c4c2ed.gif

Hence, the differential equation given in alternative D is a homogenous equation.

Also Read : Exercise-9.6-Chapter-9-Differential-Equations-class-12-ncert-solutions-Maths

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